All categories

3 years ago

A race car has a maximum speed of 0.104 km/s .What is this speed in miles per hour ?

Physics

1 answer:

sweet [91]3 years ago

8 0

Answer:

232.641374 mph

Explanation:

A race car has a maximum speed of 0.104km/s

Let X represent the speed in miles per hour

Therefore the speed in miles per hour can be calculated as follows

1 km/s = 2,236.936292 mph

0.104km/s = X

X = 0.104 × 2,236.936292

X = 232.641374

Hence the speed in miles per hour is 232.641374 mph

You might be interested in

How much does a bowling ball weigh if it has a mass of 6 kg on Earth?

larisa [96]

Answer: pretty sure 14 pounds

Explanation:

7 0

3 years ago

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)

You calculate the total work:

$W_T=Fd-F_fd=(F-F_f)d$ (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

$F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N$

Next, you replace the values of all parameters in the equation (4):

$W_T=(130N-105.84N)(5.00m)=120.80J$

$\Delta K=120.80J$

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

$W_T=\frac{1}{2}M(v_2^2-v_1^2)$ (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

$v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}$

The final speed of the box is 2.58m/s

5 0

3 years ago

slader) Two charges are arranged at corners of a square which has a side length of L = 0.25 m. The values of the charges are q1

nataly862011 [7]

Explanation:

Formula to calculate the electric potential is as follows.

$V_{A} = \frac{kq_{1}}{\sqrt{2}L} + \frac{kq_{2}}{L}$

Putting the given values into the above formula as follows.

$V_{A} = \frac{kq_{1}}{\sqrt{2}L} + \frac{kq_{2}}{L}$

= $\frac{9 \times 10^{9}}{0.25}[\frac{-3.3}{\sqrt{2}} + 4.2] \times 10^{-6}$

= $6.72 \times 10^{4} V$

Hence, electric potential at point A is $6.72 \times 10^{4} V$ .

Now, the electric potential at point B is as follows.

$V_{B} = \frac{kq_{1}}{L} + \frac{kq_{2}}{\sqrt{2}L}$

= $\frac{9 \times 10^{9}}{0.25} [-3.3 + \frac{4.2}{\sqrt{2}}] \times 10^{-6}$

= $-1.19 \times 10^{4} V$

Hence, electric potential at point B is $-1.19 \times 10^{4} V$ .

8 0

4 years ago

A girl coasts down a hill on a sled, reaching level ground at the bottom with a speed of 7.1 m/s. The coefficient of kinetic fri

Lelu [443]

Answer:

114.19186 m

Explanation:

v = Velocity of girl at bottom = 7.1 m/s

$\mu$ = Coefficient of kinetic friction = 0.045

g = Acceleration due to gravity = 9.81 m/s²

d = Distance

N = Normal force = 783 N

The weight of the sled and girl is considered as the sum of the frictional force and weight

Hence we use the following equation where the kinetic energy and potential energy are conserved

$\frac{1}{2}mv^2=\mu Nd\\\Rightarrow \frac{1}{2}\frac{783}{9.81}\times 7.1^2=0.045\times 783\times d\\\Rightarrow d=\frac{1}{2}\times \frac{783\times 7.1^2}{9.81\times 0.045\times 783}\\\Rightarrow d=114.19186\ m$

The sled travels 114.19186 m on the level ground before coming to a rest

5 0

3 years ago

Meeta used an elastic tape to measure the length of her window to stitch a curtain. Do you think she will be able to stitch a cu

Yuki888 [10]

Answer:

Explanation:

She will not be able to measure the length of her window accurately due to instrumental error from her choice of instrument. The elastic nature of her tape would alter the measurement because it will stretch as she is taking her readings, thus reducing the true measurement of the length of her window.

To measure the length of her window, she could use an inelastic tape rule or a metre rule. These instruments would eliminate instrumental error.

3 0

3 years ago