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4 years ago

5

A rollercoaster car is moving 19.8 m/s on flat ground when it hits the brakes. It accelerates at -3.77 m/s over the next 45.8 m.

How much time does it take

1 answer:

Marta_Voda [28]4 years ago

3 0

Answer:

Don't know what to do

Explanation:

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A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull

sdas [7]

In spring mass system we know that angular frequency is given as

$\omega = 2\pi f$

f = 8.38 Hz

$\omega = 2\pi(8.38)$

$\omega = 52.65 rad/s$

now we know that speed of SHM at its extreme position is given by

$v = A\omega$

here we know that

A = 17.5 cm

$v = 0.175 (52.65)$

$v = 9.21 m/s$

so maximum speed is 9.21 m/s

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3 years ago

Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that

elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p = 9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0

4 years ago

Mary pushes a crate by applying force of 18 newtons. Unable to push it alone, she gets help from her friend, Anne. Together they

maw [93]

<span>This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is:

Ff = UsN

where Us is the coefficient of static friction and N is the normal force.

In order to get the crate moving you must first apply enough force to overcome the static friction:

Fapplied = Ff

Since Fapplied = 43 Newtons:

Fapplied = Ff = 43 = UsN

and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11

43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2</span>

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3 years ago

Read 2 more answers

When atoms are split, they release energy. this concept applies to

balu736 [363]

This applies to nuclear reactions, specifically nuclear fission.

This huge release of energy has been used in atomic bombs and in the nuclear reactors that generate electricity.

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4 years ago

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

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3 years ago