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4 years ago

If a car accelerates at a constant rate of 3m/s^2, what is the velocity of the car after 4 seconds

Physics

2 answers:

vaieri [72.5K]4 years ago

6 0

It'll be 12 m/s greater than it was when the acceleration began.

Ne4ueva [31]4 years ago

3 0

V=d/t

V=3/4

V=0.75

Hope this helps and is correct :)

You might be interested in

A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is

aniked [119]

1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

$a = -\omega^2 A$

Where,

a = Acceleration

A = Amplitude

$\omega$ = Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

$a = -g$

$-\omega^2 A = -g$

$\omega = \sqrt{\frac{g}{A}}$

From the definition of frequency and angular velocity we have to

$\omega = 2\pi f$

$f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}$

$f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}$

$f = 2.5Hz$

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

6 0

3 years ago

Which statement is true for a car that first goes around a curve of radius r at a constant speed v, and then goes around the sam

sasho [114]

Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around

3 0

3 years ago

If the mass of an object increases by a factor 2, kinetic energy?

emmainna [20.7K]

Answer:

A) Increases by a factor of 2

Explanation:

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

$K.E = \frac{1}{2}MV^{2}$

Where;

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Given that mass, m = 2m

Substituting into the equation, we have;

K.E = ½mv²

K.E = ½*2mv²

Cross-multiplying, we have;

2K.E = 2mv²

Hence, if the mass of an object increases by a factor 2, kinetic energy is increased by a factor of 2.

5 0

3 years ago

0. A 3.00-kg block is dropped from rest on a vertical spring whose spring constant is 750 N/m. The block hits the spring, compre

Dmitry_Shevchenko [17]

Answer:

the spring compressed is 0.1878 m

Explanation:

Given data

mass = 3 kg

spring constant k = 750 N/m

vertical distance h = 0.45

to find out

How far is the spring compressed

solution

we will apply here law of mass of conservation

i.e

gravitational potential energy loss = gain of eastic potential energy of spring

so we say m×g×h = 1/2× k × e²

so e² = 2×m×g×h / k

we put all value here

e² = 2×m×g×h / k

e² = 2×3×9.81×0.45 / 750

e² = 0.0353

e = 0.1878 m

so the spring compressed is 0.1878 m

5 0

3 years ago

A 1.00 kg object moving in the + x direction at 10.0 m/s collides with a 1.50 kg object traveling at 5.00 m/s in the - x directi

marysya [2.9K]

Answer:

The amount of kinetic energy lost during the collision is 60.75 J

Explanation:

The given parameters are;

The mass of the object moving in the +x direction, m₁ = 1.00 kg

The velocity of the object moving in the +x direction, v₁ = 10.0 m/s

The mass of the object moving in the -x direction, m₂ = 1.50 kg

The velocity of the object moving in the +x direction, v₂ = 5.00 m/s

The final velocity of the 1.00 kg mass after collision, v₃ = 4.00 m/s

The direction of motion of the 1.00 kg mass after collision = -x direction

The total initial kinetic energy of the system, $K.E._{total \ initial}$ = 1/2·m₁·v₁² + 1/2·m₂·v₂²

∴ $K.E._{total \ initial}$ = 1/2×1.00×10² + 1/2×1.50×5² = 68.75 J

The final kinetic energy of the system, $K.E._{final}$ = 1/2·m₁·v₃²

$K.E._{final}$ = 1/2×1.00×4² = 8 J

The amount of kinetic energy lost during the collision, $\Delta K.E._{system}$ is given as follows;

$\Delta K.E._{system}$ = $K.E._{final}$ - $K.E._{total \ initial}$ = 8 J - 68.75 J = -60.75 J

The amount of kinetic energy lost during the collision = 60.75 J.

7 0

3 years ago