Ask question

Ask question

All categories

3 years ago

11

The average person passes out at an acceleration of 7g (that is, seven times the gravitational acceleration on earth). suppose a

car is designed to accelerate at this rate. how much time would be required for the car to accelerate from rest to 62.6 miles per hour? (the car would need rocket boosters!)

1 answer:

Anvisha [2.4K]3 years ago

3 0

<span>1g=32 ft per second ^2 62.6=330528ft 62.6 miles per hour =330528ft perhour=(330528ft perhour/3600seconds per hour)=91.8ft per second formula for the velocity (v(t))in feet per second at time t in seconds : v(t))=224t+v0,v0=initial velocity = 0ft per seconds v(t))=224t 224t =91.8 t=0.409seconds</span>

You might be interested in

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

2 years ago

If 30.45 grams of water is to be heated up 3.3 degrees to make baby

KengaRu [80]

Answer:

A. 420 J

Explanation:

Given the following data;

Mass = 30.45 g

Specific heat capacity = 4.18 J/g °C.

Temperature = 3.3°C

To find the quantity of heat;

Heat capacity is given by the formula;

$Q = mct$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

t represents the temperature.

Substituting into the equation, we have;

$Q = 30.45 * 4.18 * 3.3$

Q = 420.03 ≈ 420 Joules

8 0

2 years ago

When light behaves like a particle, it is called a ???

nirvana33 [79]

It is called a photon i believe

7 0

3 years ago

Read 2 more answers

An example of a household appliance with a low and high power rating

olchik [2.2K]

Answer:

Explanation:

There are countless household appliances in every single house. One appliance with a low power rating would be a ceiling fan. On average ceiling fans consume roughly 60w and are found in the majority of houses. On the other side of the spectrum, we have a high power-rating appliance such as a microwave. Microwaves use anywhere between 1000w to 1800w of power in order to function correctly. This is very large power consumption and one of the highest power ratings found for appliances in a household.

5 0

2 years ago

A 300 gg ball on a 70-cmcm-long string is swung in a vertical circle about a point 200 cmcm above the floor. The string suddenly

Ratling [72]

Answer:

the tension in the string an instant before it broke = 34 N

Explanation:

Given that :

mass of the ball m = 300 g = 0.300 kg

length of the string r = 70 cm = 0.7 m

At highest point, law of conservation of energy can be expressed as :

$\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \ m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\$

$v = 8.854 \ m/s$

The tension in the string is:

$T = \frac{mv^2}{r}\\\\T = \frac{(0.300 \ kg)*(8.854 \ m/s^2)}{0.70 \ m}\\\\T = 33.59 N\\\\T = 34 \ N$

Thus, the tension in the string an instant before it broke = 34 N

6 0

2 years ago