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2 years ago

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If a 1,000-kg car is traveling at 20.0 m/s when it reaches the bottom of a circular hill of radius 8.00 m, what normal force doe

s the road exert on the car at the bottom of the hill

1 answer:

scoray [572]2 years ago

6 0

Answer:

i don't know this one.

Explanation:

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Schach [20]

Answer:

c

Explanation:

7 0

2 years ago

Read 2 more answers

An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a

SCORPION-xisa [38]

Answer:

$a_y=0.92m/s^2$

Explanation:

To solve this problem we use the formula for accelerated motion:

$y=y_0+v_{y0}t+\frac{a_yt^2}{2}$

We will take the initial position as our reference ( $y_0=0m$ ) and the downward direction as positive. Since the rock departs from rest we have:

$y=\frac{a_yt^2}{2}$

Which means our acceleration would be:

$a_y=\frac{2y}{t^2}$

Using our values:

$a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2$

5 0

2 years ago

How are the magnetic domains of a magnet different from the domains of an ordinary piece of metal?

My name is Ann [436]

Answer:

In a magnet, the domains all point in the same direction; in an ordinary piece of metal, they're all jumbled up.

Explanation:

In a magnet, the domains all point toward the north pole; in an ordinary piece of metal, they all point to the south pole.

Side note:
Hope this helps!
Please give Brainliest!

6 0

1 year ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

2 years ago

Light passes through a pair of narrow slits with a 0.67-mm separation. It is found that the fourth bright fringe makes an angle

babunello [35]

Answer:

The wavelength of the light is 555 nm.

Explanation:

according to Bragg's law..

n×λ = d×sin(θ)

n is the fringe number

λ is the wavelength of the light

d is the slit separation

θ is the angle the light makes with the normal at the fringe.

7 0

2 years ago