Answer: C
Explanation: Some varieties of mushrooms are poisonous or may make you ill.
How do I solve this problem? String one is 5kg String 2 is 11 kg
ANSWERS
• Forces on the 5kg block: ,3
,• Forces on the 11kg block: ,2
,• Forces balanced? ,Yes, the system is at rest.
,• Tension in string 1:
,• Tension in string 2: ,107.91 N
EXPLANATION
First we have to draw the forces on each block:
Hence, block 1 has 3 forces acting on it, while block 2 has 2 forces acting on it.
It is said that the system is at rest, which means that the forces on the system are balanced.
By Newton's second law we know that,
[tex]F_{t1}-F_{t2}-F_{g1}=0[/tex]And,
[tex]F_{t2}-F_{g2}=0[/tex]Both equal zero because the blocks are not moving, and therefore there's no acceleration.
From the second equation we can find the tension in string 2,
[tex]\begin{gathered} F_{t2}=F_{g2} \\ F_{t2}=m_2\cdot g \end{gathered}[/tex]m2 = 11kg and g = 9.81m/s²,
[tex]F_{t2}=11\operatorname{kg}\cdot9.81m/s^2=107.91N[/tex]Now that we know that the tension in string 2 is 107.91N, we can find the tension in string 1 replacing this into the first equation and solving for Ft1,
[tex]F_{t1}=F_{t2}+F_{g1}[/tex][tex]undefined[/tex]URGENT!! ILL GIVE
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Answer:
Its the fourth/last answer
Need to solve step by step this exercise about fluids..
Given:
The pressure in the lower pipe is,
[tex]P_1=120\text{ kPa}[/tex]The speed of water is,
[tex]v_1=1\text{ m/s}[/tex]The radius of the lower pipe is
[tex]\begin{gathered} r_1=12\text{ cm} \\ =0.12\text{ m} \end{gathered}[/tex]the radius of the upper pipe is
[tex]\begin{gathered} r_2=6\text{ cm} \\ =0.06\text{ m} \end{gathered}[/tex]The height of the upper pipe is,
[tex]h_2=2\text{ m}[/tex]The density of water is,
[tex]\rho=1000kg/m^3[/tex]Using the continuity equation,
[tex]\begin{gathered} \pi r^2_{1^{}}v_1=\pir_2^{}^2_{}v_2 \\ v_2=\frac{r^2_{1^{}}v_1}{r^2_{2^{}}} \\ v_2=\frac{0.12\times0.12\times1}{0.06\times0.06} \\ v_2=4\text{ m/s} \end{gathered}[/tex]Hence the speed of water is 4 m/s.
Applying Bernoulli's principle we get,
[tex]\begin{gathered} P_1+\frac{1}{2}\rho(v_1)^2+\rho g\times0_{}=P_2+\frac{1}{2}\rho(v_2)^2+\rhogh_2 \\ 120\times10^3+\frac{1}{2}\times1000\times1^2+0=P_2+\frac{1}{2}\times1000\times4^2+1000\times9.8\times2 \\ P_2=120500-27600 \\ P_2=92.9\text{ kPa} \end{gathered}[/tex]Hence, the pressure is 92.9 kPa.
A 150 N crate is being pulled up a perfectly smooth ramp that slopes upward at 15 degrees by a pull that is directly at 30 degrees above the surface of the ramp. What is the magnitude of the pull required to make the crate move up the ramp at a constant velocity of 1.75 m/s?
The magnitude of the pull required to make the crate move up the ramp at a constant velocity of 1.75 m/s is 44.83 N.
What is the magnitude of the force required?
The magnitude of the force pull required to make the crate move up the ramp at a constant velocity of 1.75 m/s is calculated by applying Newton's second law of motion.
F(net) = ma
where;
F(net) is the net force on the cratem is the mass of the cratea is the acceleration of the crateat constant velocity, the acceleration of the crate = 0
F(net) = 0
Fx - Fgx = 0
where;
Fx is the horizontal component of the applied forceFgx is the horizontal component of the weight of the crateFcos(30) - 150sin(15) = 0
F = 150 sin(15) / cos30
F = 44.83 N
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Which of the following choices correctly ranks the colors of visible light from the lowest to the highest frequency?Select one:a. Ab. Bc. Cd. D
d.D
Explanation
Visible light is the small part of the electromagnetic spectrum that we can see.
Colors exist at different wavelengths from lowest energy to highest energy:
The more energy a wave has, the higher its frequency, and vice versa. When it comes to visible light, the highest frequency color, which is violet, also has the most energy. The lowest frequency of visible light, which is red, has the least energy
therefore, the visible colors from the lowest to the highest frequency are
therefore, the answer is
d.D
I hope this helps you
If a planet was located approximately 25 thousand light-years from the center of a galaxy and orbits that center once every 247 million years, how fast is the planet traveling around the galaxy in km/hr? If needed, use 3.0 × 10^8 m/s for the speed of light.
The distance of the planet from the centre i.e. radius,
[tex]r=25000\text{ light years}[/tex]The time period is
[tex]T=247\text{ million years}[/tex]The speed of the planet is given by the formula
[tex]v=\frac{2\pi r}{T}[/tex]Substituting the values, the speed will be
[tex]\begin{gathered} v=\frac{2\times\pi\times25000\text{ }\times9.461\times10^{12}}{247\times10^6\times876} \\ =6.866\times10^5\text{ km/hr} \end{gathered}[/tex]Find the power output of the engine of a 1200 kg car while the car accelerates from 30 km/h to 100 km/h in 10 s.
According to work energy thereom, the work done on the car to accelerate is,
[tex]W=\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]Plug in the known values,
[tex]\begin{gathered} W=\frac{1}{2}(1200kg)(100km/h)^2(\frac{1000\text{ m}}{1\text{ km}})^2(\frac{1\text{ h}}{3600\text{ s}})^2(\frac{1\text{ J}}{1kgm^2s^{-2}})-_{}\frac{1}{2}(1200kg)(30km/h)^2(\frac{1000\text{ m}}{1\text{ km}})^2(\frac{1\text{ h}}{3600\text{ s}})^2(\frac{1\text{ J}}{1kgm^2s^{-2}}) \\ =462963\text{ J-}41667\text{ J} \\ =421296\text{ J} \end{gathered}[/tex]The power output of the engine can be given as,
[tex]P=\frac{W}{t}[/tex]Substitute the values,
[tex]\begin{gathered} P=\frac{421296\text{ J}}{10\text{ s}}(\frac{1\text{ W}}{1\text{ J/s}}) \\ =42129.6\text{ W} \end{gathered}[/tex]Therefore, the power output of the engine is 42129.6 W.
A simple machine does 30 J of work with an efficiency of 28%. How much energy was put into the machine?
Take into account that efficiency is given by the following expression:
[tex]\text{ efficiency=(output work/input work)}\cdot100[/tex]In this case, you have:
efficieny = 28%
output work = 30J
input work = ?
Replace the previous values of the parameters into the formula for efficieny, solve for input work and simplify:
28% = (30J/input work)*100%
input work = 30J*(100% / 28%)
input work = 107.14 J
Hence,about 107.14 J was put into the machine
What kinds of nutrients would you expect to find in a slice of pizza?
please help me im stuck on the science.
Answer:
Pizza contains a mix of different nutrients such as, protein, carbohydrates, fats and calcium.
Answer:
Pizza contains a mix of nutrients such as protein, carbohydrates, fats and calcium. Grain (sometimes, but usually not, whole grain) is one of the primary ingredients in the crust.
Find the net force on the 4th part. Given, the value of velocity remain same at every point.
In the fourth case, when the weight and the tension are acting vertically downward.
The net force acting on the bucket is,
[tex]F_{\text{net}}=\frac{mv^2}{r}[/tex]where v is the velocity, r is the radius, and m is the mass,
The net force acting on the bucket will remain the same if the magnitude of the velocity remains the same.
As the direction of the bucket changes thus, the value of net force is not equal to zero.
As the velocity of the bucket is 3 m/s.
Thus, the value of net force acting on the bucket is,
[tex]\begin{gathered} F_{\text{net}}=\frac{0.5\times3^2}{0.6} \\ F_{\text{net}}=7.5\text{ N} \end{gathered}[/tex]Thus, (at the constant velocity at every point case) the net force acting is 7.5 N.
A person is pushing on a car with a force of <-14,4> and another person is pushing on a car with a force of -6,1> What is the net force in the vertical direction?
We are given the following vector forces:
[tex]\begin{gathered} F_1=(-14,4) \\ F_2=(-6,1) \end{gathered}[/tex]These vector forces are expressed as:
[tex]F=(x,y)[/tex]Where the first component "x" is the horizontal component, and "y" is the vertical component. Since we are asked about the net force in the vertical component we need to add the components "y" of both vectors, we get:
[tex]\Sigma F_y=4+1=5[/tex]Therefore, the net force in the vertical direction is 5.
is 11/12 greater = or less then 2/3
11/12 is greater than 2/3
Explanation
to solve this we have two options, rewriting them in terms of a common denominator, or the fastest way to compare fractions is to convert them into decimal numbers.
Step 1
a)rewriting in terms of a common factor
we can see that the denominator of the second fraction is 3, so we can multiply by (4/4) in order to ger 12 as denominator , so
[tex]\frac{2}{3}\cdot\frac{4}{4}=\frac{8}{12}[/tex]note that we are not changing or affecting the fraction ( we are just multipliyng by 1)
Step 2
now, we have both fraction with the same denominator, let's compare
When the denominators are the same, the fraction with the lesser numerator is the lesser fraction and the fraction with the greater numerator is the greater fraction
[tex]\frac{11}{12}\text{ and }\frac{8}{12}[/tex]therefore, the greater fraction is 11/12
in other words
11/12 is greater than 2/3
I hope this helps you
A circuit breaker is a short piece of metal that melts when a current that is too large passes through it. Is this true or false?
A circuit beaker is a short piece of metal that melts when a current that is too large passes through it. This given statement is false statement.
A block weighing 200N is pushed along a surface. If it takes 80N toget the block moving and 40N to keep the block moving at a constantvelocity, what are the coefficient of friction us and uk ?
Answer:
see below
Explanation:
coefficient of static F = 80 / 200 = .4
kinetic F = 40 /200 = .2
rock that is attached to a 2.6 m rope is whirled around and covers 3 Radians every 1.2seconds. What is the Tangential velocity of the rock when the rope is released?
Answer: Tangential velocity = 6.5 m/s
Explanation:
The formula for calculating tangential velocity is expressed as
v = rw
where
v represents tangential velocity
r represents the radius of the circular path
w is the angular velocity
From the information given,
r = 2.6
If 3 radians is covered in 1.2 seconds, then
angular velocity, w = distance /time = 3/1.2
w = 2.5 rad/s
Thus,
v = 2.6 x 2.5
v = 6.5 m/s
Tangential velocity = 6.5 m/s
An object weighing 297 N in air is immersedin water after being tied to a string connectedto a balance. The scale now reads 263 N.Immersed in oil, the object appears to weigh270 N.Find the density of the object.Answer in units of kg/m^3Do not round.
Given
Wair = 297 N
Wwater = 263 N
Density of water = 1000 kg/m^3
Acceleration due to gravity = 9.8
Procedure
We can calculate the thrust of the water with the difference of weights
[tex]\begin{gathered} W_{\text{air}}-W_{\text{water}}=\rho_wgV \\ V=\frac{W_{\text{air}}-W_{\text{water}}}{\rho_wg} \\ V=\frac{297N-263N}{1000\cdot9.8} \\ V=0.00347\text{ m}^3 \end{gathered}[/tex]Now for the object
[tex]\begin{gathered} \rho_{\text{obj}}=\frac{W_{\text{air}}}{gV} \\ \rho_{\text{obj}}=\frac{297}{9.8\cdot0.00347} \\ \rho_{\text{obj}}=8735.29kg/m^3 \end{gathered}[/tex]
What is the potential difference across a curling iron if there is a current of 2.5 A transferring 9360 J in 32s?
Power is expressed as energy/time
Thus means that
power = energy/time
From the information given,
energy = 9360 J
time = 32 s
Thus,
power = 9360/32 = 292.5 J/s = 292.5 watts
The formula relating potential difference which is the same as voltage, current and power is expressed as
P = IV
where
P = power = 292.5
I = current = 2.5
V = potential difference = ?
Thus, we have
292.5 = 2.5V
V = 292.5/2.5
V = 117 V
The potential difference is 117 V
What is the net change in internal energy (ΔU) if the amount of work done (W) is the same as the amount of energy transferred in by heat (Q)?
ANSWER:
Zero
STEP-BY-STEP EXPLANATION:
Heat (Q) and work (W) are the two ways to add or remove energy from a system. The processes are very different, however, both can change the internal energy (ΔU) of a system.
It is given by the following equation:
[tex]\begin{gathered} \Delta U=Q-W \\ \\ \text{ So if they are the same:} \\ \\ Q=W \\ \\ \Delta U=Q-Q \\ \\ \Delta U=0 \end{gathered}[/tex]The net change in internal energy (ΔU) is zero
A wire with resistance of 8.0 Ω is drawn out through a die such that its new length is twice its original length. Find the resistance of the longer wire assuming that the resistivity and the density of the material of the wire are unchanged during the drawing process.
Explanation
the electrical resistivity is define as
[tex]\begin{gathered} \sigma=R\frac{A}{l} \\ where \\ R\text{ is the electrical resitance} \\ \text{A is the cross seccional area} \\ l\text{ is the length} \end{gathered}[/tex]so,if we isolate R
[tex]\begin{gathered} \sigma= R\frac{A}{l} \\ R=\sigma\frac{l}{A} \end{gathered}[/tex]hence, the ratio of the resitances is
[tex]\frac{R_1}{R_2}=\frac{\sigma\frac{l}{A}}{\sigma\frac{l_2}{A}}[/tex]the volume of the wire is constant , therefore
[tex]A_1l_1=A_2l_2[/tex]if the new length is twice the original
The resistance of the longer wire assuming that the resistivity and the density of the material of the wire are unchanged during the drawing process is twice the resistance of the wire before, i.e. 16 Ω.
What is Resistance?The impediment to current flow in an electrical circuit is measured by resistance. The ohm, a unit of measurement for resistance, is represented by the Greek letter omega. The name of the unit of resistance is derived from Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current, and resistance.
Given:
The resistance of the wire, R = 8 Ω.
Calculate the electrical resistivity by the formula given below,
σ = [tex]R\frac{A}{l}[/tex],
Here σ is the electrical resistivity, A is the area, and l is the length of the wire.
According to Que's statement,
R₁ / R₂ = {σ₁ l₁ / A₁ } / {σ₂l₂ / A₂}
R₂ = 2 *R₁
R₂ = 2*8
R₂ = 16 Ω
Therefore, the resistance of the longer wire assuming that the resistivity and the density of the material of the wire are unchanged during the drawing process is twice the resistance of the wire before, i.e. 16 Ω.
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Which pan of water shows molecules that have received the most heat from a stove?Select one:a. Ab. Bc. Cd. D
Given:
The water in the pan is heated the most.
To find:
Pan of water that has received the most heat from the stove.
Explanation:
When water is heated, the water molecules in the water acquire kinetic energy due to a rise in water temperature. The more the water is heated, the higher the temperature of the water. Thus the kinetic energy of molecules of water will be more.
The higher kinetic energy of water molecules corresponds to the higher speed of the water molecules.
Thus, in option (a), the water molecules are moving very fast, meaning they have high kinetic energy due to high temperature. And the temperature will be maximum if we provide maximum heat to the water.
Thus, option (a) is correct.
Conclusion: The correct option is option (a).
As a 600 N woman sits on the floor, the floor exerts a force on her ofa. 6 Nb. 60 NC. 1200 Nd. 600 N
the floor exerts a force of 600 N on her
Explanation
Newton's third law states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction,in this case the wowan exerts a force on the floor, it is her weigth, in the other hand, the floor exerts a force on the woman,this is called Normal force, the normal force is the force that surfaces exert to prevent solid objects from passing through each other. Normal force is a contact force.
Step 1
a) diagram:
as the woman is at equilibrium , the sum of the forces acting on it equals, zero, hence
[tex]\begin{gathered} Normal\text{ force-weigth= 0} \\ Normal\text{ force = weigth} \\ Normal\text{ force=600 N} \end{gathered}[/tex]therefore, the floor exerts a force of 600 N on her
so, the answer is
d) 600 N
I hope this helps you
An object is dropped from the top of a building. How fast is it moving after 5s? What is the acceleration at this time?
Any object under free fall accelerates at a constant rate given by the gravitational acceleration:
[tex]g=9.81\frac{m}{s^2}[/tex]On the other hand, the speed v of an object under free fall after t seconds, if it starts from rest, is given by the formula:
[tex]v=gt[/tex]Replace g=9.81m/s^2 and t=5s to find the speed of the object 5 seconds after the object is dropped:
[tex]v=(9.81\frac{m}{s^2})(5s)=49.05\frac{m}{s}[/tex]Therefore, the speed of the object after 5 seconds is approximately 49 meters per second. Its acceleration is always the same and it is equal to 9.81 meters per second squared.
3. 1.819 m4. 5.291 m5. 6.321 m39. Answer: B40. Work done by the non conservative forces actingon an object is equal1.to the change in the mechanical energy of the object2. to the change in the kinetic energy of the object3. to the work done by the conservative forces4. to the change in the potential energy of the object5. to the net work done on the object41. Answer: A
We are asked to determine the distance that spring will stretch when a given mass is attached to it.
To do that we will use Hook's law:
[tex]F=kx[/tex]where:
[tex]\begin{gathered} F=\text{ force } \\ k=\text{ spring constant} \\ x=\text{ distance that the spring is stretched} \end{gathered}[/tex]We will determine the constant of the spring "k" first using the fact that the spring is stretched 0.8 meters when a mass of 3kg hangs from it.
Since the only force acting on the spring is the weight of the object we have:
[tex]F=mg[/tex]Where:
[tex]\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now, we substitute and we get:
[tex]mg=kx[/tex]Now, we divide both sides by "x":
[tex]\frac{mg}{x}=k[/tex]Now, we plug in the values:
[tex]\frac{(3kg)(9.8\frac{m}{s^2})}{0.8m}=k[/tex]Solving the operations:
[tex]36.75\text{ N/m}=k[/tex]Now, we substitute the value of "k":
[tex]F=(36.75\text{ N/m\rparen}x[/tex]Now, we solve for "x":
[tex]\frac{F}{36.75\text{ N/m}}=x[/tex]Now, we substitute the value of the weight of the second object:
[tex]\frac{(14kg)(9.8\frac{m}{s^2})}{36.75\text{ N/m}}=x[/tex]Solving the operations:
[tex]3.733m=x[/tex]Therefore, the spring will stretch by 3.733 meters.
An air-filled pipe is found to have successive harmonics at 800 Hz , 1120 Hz , and 1440 Hz . It is unknown whether harmonics below 800 Hz and above 1440 Hz exist in the pipe. The length is 53.5 cm. Identify the correct pressure variation graph for the 1120 Hz standing wave in the pipe. Note that the closed end of the pipe is on the right.
Using the concept of Harmonic-wave, we got the desired graph which is shown in the image.
It must be an open-closed pipe.
Open-closed pipe
Let the fundamental frequency of f be
The second harmonic is 3f.
3rd harmonic = 5f
Therefore, the difference between harmonics is 3f-f = 2f
So, 2f = (1120-800) = 1440-1120 = 320
Also, f = 160Hz
800Hz = 5th Harmonic
1120 Hz = 7th
1440 = 9th
For an open- or closed-pipe,
f = v/4L
where v is the speed of sound in the air = 343 M/s
So, 160 = 343/4L
Also, L = 0.5336 m = 53.6cm
Hence for the frequency of 1120Hz, we got the desired graph of a wave which is shown in the below graph.
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instantaneous velocity for a displacement function dt=2-2t at any given time
a -2t
b-2
c 2
d 2t
Instantaneous velocity for a displacement function [tex]d_{t}[/tex] = 2-2t is -2 m/s. So the correct option is (b)
What is instantaneous velocity?Instantaneous velocity is defined as the rate of change of position over a very short (near zero) time interval. Measured in SI units m/s. The instantaneous velocity is the magnitude of the instantaneous velocity. Same value as instantaneous velocity, but without direction. Simply put, the velocity of an object at that point in time is called the instantaneous velocity. Hence the definition is given as "velocity of a moving object at a given point in time". It can also be determined by taking the slope of the displacement-time plot or the x-t plot.
If the object's velocity is constant, the instantaneous velocity can be the same as the default velocity.
For the given case,
Displacement function [tex]d_{t}[/tex] = 2-2t
Assume x = 2-2t
Time = t
Since [tex]v_{t}[/tex] = [tex]\frac{d_{} _{x} }{d_{t} }[/tex]
[tex]v_{t}[/tex] = [tex]\int\limits^t_0[/tex]2 [tex]d_{t}[/tex] - [tex]\int\limits^t_0[/tex]-2t [tex]d_{t}[/tex]
[tex]v_{t}[/tex] = -2 m/s
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if a rocket travles 5600km in 3 hours, what is its speed?
Given the distance traveled by rocket is
[tex]D=5600[/tex]The time taken by the rocket is 3 hours.
[tex]T=3\text{ hours}[/tex]Calculate the speed of the rocket.
[tex]\begin{gathered} \text{Speed = }\frac{5600km}{3\text{ hours}} \\ \text{Speed = 1866.66 }\frac{km}{hr} \end{gathered}[/tex]Thus, the time taken by the rocket is 1866.66 km/hr.
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
An ice cube is placed on a metal block that is at a room temperature of 25°C Which statement best describes the result of this interaction?
A. Heat is transferred from the ice cube to the metal block, and the ice melts
B. Heat is transferred from the air to the ice cube, and the ice melts.
C. Heat is transferred from the air and metal to the ice cube, and the ice melts
D. Heat is transferred from the ice cube to the air, and the ice melts
Answer:
C. Heat is transferred from the air and metal to the ice cube, and the ice melts
A set of charged plates have anarea of 5.10*10^-3 m^2 andseparation 1.42*10^-5 m. Howmuch charge must be placed onthe plates to create a potentialdifference of 125 V across them?(The answer is *10^-7 C. Just fill inthe number, not the power.)
Given data:
Area of plates:
[tex]A=5.10\times10^{-3}\text{ m}^2[/tex]Separation between the plates:
[tex]d=1.42\times10^{-5}\text{ m}[/tex]Potential difference:
[tex]V=125\text{ V}[/tex]The capacitance of the capacitor is given as,
[tex]C=\frac{A\epsilon_{\circ}}{d}[/tex]Here, ε_o is the permittivity of the free space.
Substituting all known values,
[tex]\begin{gathered} C=\frac{5.10\times10^{-3}\times8.85\times10^{-12}}{1.42\times10^{-5}} \\ =3.178\times10^{-9}\text{ F} \end{gathered}[/tex]The charge on the capacitor is given as,
[tex]Q=CV[/tex]Substituting all known values,
[tex]\begin{gathered} Q=3.178\times10^{-9}\times125 \\ =3.9725\times10^{-7}\text{ C} \end{gathered}[/tex]Therefore, the charge on the plates is 3.9725×10^-7 C.
From the list of world record track events in the table, calculate the average speeds for each race. Assume the length of each event is known to the nearest 0.1 m. Calculate only the last one, for 10,000 m.
Answer:
0.095 m/s
Explanation:
The average speed can be calculated as
Avg speed = distance/time
For the last one, we get that distance is 10000 m and the time 29 hours, 17 min, and 45 seconds.
So, first, we need to convert 29:17.45 to seconds as follows
29 hours x 3600 s/ 1 hour = 104400 s
17 min x 60 s / 1 min = 1020 s
45 s = 45 s
Total time = 104400 s + 1020 s + 45 s
Total time = 105,465 s
Then, the average speed is equal to
Avg speed = 10,000 m/105,465 s
Avg speed = 0.095 m/s
Therefore, the answer is 0.095 m/s
A particle moves in a straight line, and you are told that the torque acting on it is zero about some unspecified origin. Does this necessarily imply that the total force on the particle is zero? Can you conclude that its angular velocity is constant?
Horizontal force will be there but angular force is zero because it is dependent on angle and angular velocity is constant.
[tex]\tau = rF\sin\theta[/tex]
T= torque
r= radius
In mechanics, every action that tries to preserve, alter, or change a body's motion is referred to as a force. Isaac Newton's Principia Mathematica has three principles of motion that are usually used to explain the concept of force (1687).
Newton's first law states that in the absence of an external force, a body will continue to be in either its resting or evenly moving condition along a straight path. According to the second law, any time an outside force acts on a body, the body accelerates (changing velocity) in the force's direction.
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