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2 years ago

A 10 kg brick and a 1 kg book are dropped in a vacuum. The force of gravity on the 10 kg brick is what?

2 answers:

7 0

<span>10 times as much. Since F=m*a, and a is constant, the only thing that affects force is the mass.

In response to the below answer, the acceleration due to gravity does not change. The force due to gravity definitely DOES change depending on the mass of the object. Since the force is what the problem asks for, the answer is 10</span>

LuckyWell [14K]2 years ago

3 0

If this is happening on Earth, then the force of gravity between the Earth and the brick is 98 Newtons. It doesn't matter whether the brick is falling, rising, dropped in a vacuum, in water, or into chicken soup, or what else is dropped near it. None of that has any effect on the gravitational force between the Earth and the brick.

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A rock with density 1900 kg/m3 is suspended from the lower end of a light string. When the rock is in air, the tension in the st

wel

Answer:

the tension T2 when the rock is completely immersed is T2 = 29.05 N

Explanation:

from Newton's second law

F= m*a

where F= force , m= mass , a= acceleration

when the rock is suspended ,a=0 since it is at rest. Then

T1 - m*g = 0 , T1= tension when suspended in air , g= gravity

assuming constant density of the rock

m= ρ rock *V , where ρ rock = density of the rock , V= volume

thus

T1= m*g = ρ rock *g*V

V= T1/(ρ rock *g)

when the rock is submerged in oil , it receives an upward force that equals the weight of the volume of displaced oil (V displaced). Since it is completely submerged the volume displaced is the volume of the rock V=Vdisplaced

When the rock is at rest , then

F= m*a=0

T2 + ρ oil *g*V displaced - ρ rock *g*V =0

T2 = ρ rock *g*V - ρ oil *g*V = g*V (ρ rock - ρ oil)

T2 = g*V (ρ rock - ρ oil) = T1/(ρ rock *g) *g * (ρ rock - ρ oil)

T2 = T1 * (ρ rock - ρ oil)/ρ rock

replacing values

T2 = 48 N * (1900 kg/m3- 750 kg/m3)/ 1900 kg/m3 = 29.05 N

T2 = 29.05 N

3 0

2 years ago

A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force

VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0

2 years ago

Helium (density 0.18 kg/m^3 at 0°C and 1 atm pressure) remains a gas until the extraordinarily low temperature of 4.2 K.What is

balu736 [363]

Answer:

$166.6396m/sec$

Explanation:

Molar mass of helium = $4\times 10^{-3}kg/mole$

$\gamma$ for helium is 1.67

The velocity of helium in sound at any temperature is given by $v=\sqrt{\frac{\gamma RT}{M}}=\sqrt{\frac{1.67\times 8.314\times 8}{4\times 10^{-3}}}=166.6396m/sec$

R is a constant 8.314 $atm\ mol^{-1}K^{-1}$

7 0

2 years ago

An electric winch is used to raise a 40-kg package and a 10-kg package vertically up the side of a building as pictured in the d

just olya [345]

Answer:

Explanation:

40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.

3 0

2 years ago

A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the

cupoosta [38]

Answer: deceleration of $1.904\ rad/s^2$