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3 years ago

14

Your mass is 65 kg. you stand on a bathroom scale in an elevator on earth. (a) what force would the scale exert when the elevato

r moves up at a constant speed?

1 answer:

erik [133]3 years ago

4 0

I think you would use F = ma
F = 65*10
F = 650N
(The 10m/s is from acceleration due to gravity)

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Big Ben, a large artifact in England, has a mass of 1x10^8 kilograms and the Empire State Building 1x10^9 kilograms. The distanc

TiliK225 [7]

Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

The mass of Big Ben, M₁ = 1 × 10⁸ kg

The mass of the Empire State Building, M₂ = 1 × 10⁹ kg

The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

$F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}$

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

$F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}$

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

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What happens when light enters water?

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The answer should be A

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3 years ago

A 1870 kg car traveling at 13.5 m/s collides with a 2970 kg car that is initally at rest at a stoplight. The cars stick together

Inga [223]

Answer:

The value is $\mu = 0.72$

Explanation:

From the question we are told that

The mass of the first car is $m_1 = 1870\ kg$

the initial speed of the car is $u = 13.5 \ m/s$

The mass of the second car is $m_2 = 2970\ kg$

The distance move by both cars is s = 1.93 m

Generally from the law of momentum conservation

$m_1 * u_1 + m_2 * u_2 = (m_1 + m_2 ) * v_f$

Here $u_2 = 0$ because the second car is at rest

and $v_f$ is the final velocity of the the two car

So

$1870* 13.5+ 0= ( 1870 + 2970 ) * v_f$

=> $v_f = 5.22\ m/s$

Generally from kinematic equation

$v_f^2 = u_2^2 + 2as$

here a is the deceleration

So

$5.22^2 = 0 + 2 *a * 1.93$

=> $a = 7.06 \ m/s^2$

Generally the frictional force is equal to the force propelling the car , this can be mathematically represented as

$F_f = F$

Here F is mathematically represented as

$F = (m_1 + m_2) * a$

$F = (1870 + 2970) * 7.06$

$F =34170.4 \ N$

and

$F_f = \mu * (m_1 + m_2 ) * g$

$F_f = 47432 * \mu$

So

$47432 * \mu = 34170.4$

=> $47432 * \mu = 34170.4$

=> $\mu = 0.72$

6 0

3 years ago