All categories

2 years ago

A car accelerates uniformly from rest to a speed of 6.6 m/s in 6.5 s. Find the distance the car travels during this time.

Physics

2 answers:

andrew11 [14]2 years ago

6 0

It traveled 39.6 meters

stiks02 [169]2 years ago

3 0

Answer:

Distance covered by the car, s = 21.56 meters

Explanation:

Given that,

Initially, the car is at rest, u = 0

It reaches a speed, final speed, v = 6.6 m/s

Time, t = 6.5 s

We need to find the distance covered by the car during this time. Firstly, we can find the acceleration of the car using equation of motion as :

$v=u+at$

$6.6=0+a\times 6.5$

$a=1.01\ m/s^2$

Now using third equation of motion to find distance (s) as :

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{(6.6)^2}{2\times 1.01}$

s = 21.56 meters

So, the distance traveled by the car during 6.5 seconds is 21.56 meters. Hence, this is the required solution.

You might be interested in

A small lab cart and one of larger mass collide and rebound off each other. Which of them has the greater average force on it du

I am Lyosha [343]

When a small lab cart collide with a large mass then during the collision two bodies will remain in contact and then apply the contact force on each other

This contact force is always equal on two bodies in magnitude and also it is normal or perpendicular to the contact surface.

This will always follow Newton's III law as per which every action has equal and opposite reaction.

So this contact force will not depend on the mass or any other factor but it will remain same on two colliding bodies.

So here the correct answer would be

They both experience the same magnitude of the collision force.

7 0

1 year ago

Part A

irina [24]

Answer:

v' = -18 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$p_{o} = p_{f} (1)$

The initial momentum can be expressed as follows (taking as positive the initial direction of the ball):

$m_{b} * v_{b} -M_{c}*V_{c} = m_{b} * 18 m/s + (-M_{c}* 20 m/s) (2)$

The final momentum can be expressed as follows (since we know that v'b is opposite to the initial vb):

$-(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)$

If we assume that Mc >> mb, we can assume that the car doesn't change its speed at all as a result of the collision, so we can replace V'c by Vc in (3).
So, we can write again (3) as follows:

$-(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (4)$

Replacing (2) and (4) in (1), we get:

$m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (5)$

Simplifying, and rearranging, we can solve for v'b, as follows:
$v'_{b} = -18 m/s (6)$ , which is reasonable, because everything happens as if the ball had hit a wall, and the ball simply had inverted its speed after the collision.

3 0

1 year ago

A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106 hz and 1hz=

AURORKA [14]

The frequency of the radio station is
$f=88.7 fm= 88.7 MHz = 88.7 \cdot 10^6 Hz$

For radio waves (which are electromagnetic waves), the relationship between frequency f and wavelength $\lambda$ is
$\lambda= \frac{c}{f}$
where c is the speed of light. Substituting the frequency of the radio station, we find the wavelength:
$\lambda= \frac{3 \cdot 10^8 m/s}{88.7 \cdot 10^6 Hz}=3.38 m$

6 0

2 years ago

Read 2 more answers

A sky diver steps from a high-flying helicopter. if there were not air resistance, how fast would she be falling at the end of a

Rama09 [41]

If there is no air resistance and the body, in this case the sky diver simply falls from the helicopter then, the motion can be described as a free-falling body. The equation that would allow us to determine the speed of the body at any time, t is,

v(t) = vo + gt

where v(t) is the unknown velocity, vo is the initial velocity which is equal to zero and g is the acceleration due to gravity which is equal to 9.8 m/s2.

Substituting the known values,

v(t) = 0 + (9.8 m/s2)(12s)

v(t) = 117.6 m/s

Hence, the speed of the sky diver at the end of 12 second is approximately 117.6 m/s.

8 0

1 year ago

If the temperature of an ideal gas is increased from 200 K to 600 K, what happens to the rms speed of the molecules? (a) It incr

lawyer [7]

Answer: (d) It is !3 times the original speed.

Explanation: The rms speed of a gas is related to its temperature by the formulae below;

U(r.m.s) =√(3RT)/M

Where;

T represents the temperature.

R represents the gas constant.

M represents the molar mass of the gas.

Therefore, if the temperature increases from 200k to 600k

The temperature has then increased by a factor of 3,

However, we must note that temperature in the formulae is included in the square-root

Recall,

U(r.m.s) =√(3RT)/M

Consequently, temperature (T) can now be represented by (3T).

The inference drawn from this is that the root-mean-square speed would increase by a factor of √3

Therefore, option (d) is correct.

7 0

2 years ago

Read 2 more answers