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3 years ago

Where do you weight more on top of Mt. Everest (Highest Mt. in the world) or at the beach

1 answer:

8 0

You weigh slightly more at the beach than on top of a mountain, because at the beach, you're closer to the center of the Earth.

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A uniformly charged ring of radius 10.0 cm has a total charge of 50.0 μC Find the electric field on the axis of the ring at 30.0

Grace [21]

Answer: 4.27 *10^6 N/C

Explanation: In order to calculate the electric field along the axis of charged ring we have to use the following expression:

E=k*x/(a^2+x^2)^3/2 where a is the ring radius and x the distance to the point measured from the center of the ring.

Replacing the data we have:

E= (9* 10^9* 0.3* 50 * 10^-6)/(0.1^2+0.3^2)^3/2

then

E=4.27 * 10^6 N/C

8 0

3 years ago

Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0

4 years ago

You are In-line skates at the top of a small hill. Your potential energy is equal to 1000 J. The last time we checked, your mass

stira [4]

A) Weight = mass × acceleration (due to gravity)

= 60×9.8

= 588 N

B) Potential energy = mass x gravity x change in height

1,000 = 60.0 x 9.8 x h

h = 1.7 m

C) Kinetic energyF = potential energyI

KEF = 1/2mv2

PEI = mgh = 1,000 J

1/2mv2 = 1,000

1/2(60.0)v2 = 1,000

v2 = 33.33

v = 5.77 m/s

3 0

4 years ago

What is denser medium?

abruzzese [7]

Answer: A medium in which speed of light is more is known as optically rarer medium and a medium in which speed of light is less is said to be optically denser medium. For example in air and water, air is raer and water is a denser medium.

Explanation:

6 0

3 years ago

Một cần trục có trọng lượng Q = 50 kN cẩu vật nặng có trọng lượng P = 10 kN

lesantik [10]

3373727227717177

Explanation:

yun hehew

8 0

3 years ago

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