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2 years ago

6

a car takes 125 ft to brake from 60 to 0 mph. Assume that the acceleration of the Prius is constant while braking. Find how long

it takes for the Prius to come to a stop. Long problems begin on the next page

1 answer:

max2010maxim [7]2 years ago

7 0

Answer:

2.84 seconds

Explanation:

t = ?

distance = 125

Velocity origianal = 60 m/hr = 88 ft/s

AVERAGE velocity = 88/2 = 44 ft/s

44 t = 125

t = 125/44 = 2.84 s

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Which of the following are true for E = mc2?

lozanna [386]

The question is very poor.

The answer it's fishing for is choice (b), but the only way to know that is
to observe that the others are even worse, and totally from outer space.

This is a Physics question, for heaven's sake. Real science is built on
measurements, and nothing gets done by operating with quantities like
a "small" change in mass, or a "great deal" of energy. This question
causes me to shiver with indignation.

6 0

3 years ago

A toy rocket, mass 0.8 kg, blasts up at a 45 degrees from ground level with a kinetic energy of 41 J. To what maximum vertical h

wel

Answer:

The maximum height reached by the toy rocket is 0.73 meters.

Explanation:

It is given that,

Mass of the toy rocket, m = 0.8 kg

It is projected at an angle of 45 degrees from ground level. The kinetic energy of the rocket is used to find the velocity with which it was projected as :

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

$v=\sqrt{\dfrac{2K}{m}}$

$v=\sqrt{\dfrac{2\times 41}{0.8}}$

u = 10.12 m/s

When it reaches to a maximum height, its final velocity will be equal to 0, v = 0

Using third equation of motion to find it as :

$v^2-u^2=2as$

a = -g

$-u^2=-2gs$

s is the maximum height reached by the toy rocket

In vertical direction, $u_y=u\ sin\theta$

$s=\dfrac{u_y}{2g}$

$s=\dfrac{u\ sin\theta}{2g}$

$s=\dfrac{10.12\times sin(45)}{2\times 9.8}$

s = 0.36 meters

So, the maximum height reached by the toy rocket is 0.73 meters. Hence, this is the required solution.

4 0

3 years ago

Particles q1 = -53.0 uc, q2 = +105 uc, and

Nimfa-mama [501]

Answer:

$-180.38\ \text{N}$

Explanation:

$q_1=-53\ \mu\text{C}$

$q_2=105\ \mu\text{C}$

$q_3=-88\ \mu\text{C}$

r = Distance between the charges

$r_{12}=0.5\ \text{m}$

$r_{23}=0.95\ \text{m}$

$r_{13}=1.45\ \text{m}$

k = Coulomb constant = $9\times 10^9\ \text{Nm}^2/\text{C}^2$

Net force is given by

$F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}$

The force on the particle $q_1$ is $-180.38\ \text{N}$ .

8 0

3 years ago

A good model of the cell membrane would be

Nonamiya [84]

B. A film of food wrap. The cell membrane is the outermost layer that is protecting the cell

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3 years ago

A toy car is moving across a table with a velocity of 7.0 m/s and drives off the end. The table is 1.8 m tall. How long will the

KatRina [158]

Answer:

Option D - 0.2 s

Explanation:

We are given;

Initial velocity; u = 7 m/s

Height of table; h = 1.8m

Now,since we want to find the time the car spent in the air, we will simply use one of Newton's equation of motion.

Thus;

h = ut + ½gt²

Plugging in the relevant values, we have;

1.8 = 7t + ½(9.8)t²

4.9t² + 7t - 1.8 = 0

Using quadratic formula to find the roots of the equation gives us;

t = -1.65 or 0.22

We can't have negative t value, thus we will pick the positive one.

So, t = 0.22 s

This is approximately 0.2 s

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3 years ago