A car started from a rest and accelerated at 9.54 m/s2 for 6.5 seconds. How much distance was covered by the car?

Super Mario and Bowser Jr. are racing around a track when Baby Bowser launches a green shell at Mario, bringing him to rest. Bow

prisoha [69]

Answer:

(a). Mario's speed in m/s = 5.2 × 10^-3 m/s.

Bowser Jr.'s speeds in m/s = 3.92 × 10^-3 m/s.

(b). 27001.2 seconds(s)..

(c). 141 metre(m).

Explanation:

So, the following data or parameters or information was given in the question above. These informations are going to help us in solving this question or problem;

=>" Bowser Jr. then passes Mario at his top speed = 30 blocks/h.

=> " Mario quickly accelerates and reaches his top speed of 40 blocks/h in order to catch back up and pass Bowser Jr., but by this point he has opened up a 75 block head start down the track."

=> "Mario world measure distance using the units of blocks, with 1 block = 0.47 m"

Therefore, the solution is given below;

(1). For the first part, we are to determine or calculate Mario and Bowser Jr.'s speeds in m/s.

Therefore, Mario's speed in m/s = 40 × 0.47) ÷ 3600 = 5.2 × 10^-3 m/s.

Also, Bowser Jr.'s speeds in m/s = ( 30 × 0.47) ÷ 3600 = 3.92 × 10^3 m/s.

(2). So, the next thing to do now.is to determine or calculate how long it took for Mario to catch Bowser Jr.

Thus, the time it took for Mario to catch Bowser Jr. Can be related as below;

[ ( 5.2 × 10^-3 m/s) - (3.92 × 10^-3 m/s) × (time,t taken for Mario to catch Bowser Jr.) = 75 × 0.47.

Therefore, the time it took for Mario to catch Bowser Jr. = 27001.2 seconds.

(3). Now, we calculate How far down the track Mario from the point at which he reaches his top speed.

The distance = 5.2 × 10^-3 m/s × 27001.2m = 141m

6 0

3 years ago

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the stri

Dmitrij [34]

Answer:

(a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

Explanation:

Given that,

Radius = 8.00 cm

Mass = 0.180 kg

Height = 75.0 m

We need to calculate the angular speed of the rotating

Using conservation of energy

$\dfrac{1}{2}I\omega_{1}^2+\dfrac{1}{2}mv_{1}^{2}+mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

Here, initial velocity and angular velocity are equal to zero.

$mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

$mg(h_{1}-h_{2})=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}$

$mgH=\dfrac{1}{2}mr^2\omega_{2}^2+\dfrac{1}{2}m(r\omega_{2})^2$

Here, $H = h_{1}-h_{2}$

$gH=r^2\omega_{2}^2$

$\omega_{2}^2=\dfrac{gH}{r^2}$

$\omega_{2}=\sqrt{\dfrac{9.8\times75.0\times10^{-2}}{(8.00\times10^{-2})^2}}$

$\omega_{2}=33.8\ rad/s$

The angular speed of the rotating is 33.8 rad/s.

(b). We need to calculate the speed of its center

Using formula of speed

$v=r\omega$

Put the value into the formula

$v=8.00\times10^{-2}\times33.8$

$v=2.7\ m/s$

Hence, (a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

3 0

3 years ago

Read 2 more answers

On what does kinetic energy depend?

tekilochka [14]

Answer:

b

Explanation:

7 0

2 years ago

Read 2 more answers

At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is

Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

$\dfrac{1}{2}mv^2=mgl$