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3 years ago

A 4900 n gear box with a mass of 500kg falls to the ground. What is it’s acceleration due to gravity?

Physics

1 answer:

Lena [83]3 years ago

8 0

Answer:

$9.8 m/s^2$ , towards the ground

Explanation:

For an object in free fall, the only force acting on it is the force of gravity. We can write it as

$F=mg$

where

F is called weight of the object

m is the mass of the object

g is the acceleration due to gravity

We can re-arrange the equation as

$g=\frac{F}{m}$

For the box in the problem, we know:

F = 4900 N

m = 500 kg

Therefore, we can find the value of g:

$g=\frac{4900}{500}=9.8 m/s^2$

And the direction is the same as the force of gravity (downward).

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A stretched rubber band is an example of which type potential energy

Alenkinab [10]

Answer: elastic potential energy

Explanation:

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2 years ago

Read 2 more answers

The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.

Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let $\rho_{m}$ be the density of metal and $\rho_{w}$ be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

$W =0.10400\ N$

$mg=0.10400$

$\rho V g=0.10400$

$Vg=\dfrac{0.10400}{\rho_{m}}$ .....(I)

The weight of metal in water is

Using buoyancy force

$F_{b}=0.10400-0.08400$

$F_{b}=0.02\ N$

We know that,

$F_{b}=\rho_{w} V g$ ....(I)

Put the value of $F_{b}$ in equation (I)

$\rho_{w} Vg=0.02$

Put the value of Vg in equation (II)

$\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02$

$1000\times\dfrac{0.10400}{0.02}=\rho_{m}$

$\rho_{m}=5200\ kg/m^3$

Hence, The density of the metal is 5200 kg/m³.

6 0

3 years ago

An asteroid is on a collision course with Earth. An astronaut lands on the rock to bury explosive charges that will blow the ast

forsale [732]

Answer:

The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters

Explanation:

Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;

v² = u² - 2·g·h

Where;

v = The final velocity

u = The initial velocity

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height she jumps

At the maximum height, $h_{max}$ = 0.500 m, she jumps, v = 0, therefore, we have;

0² = u² - 2·g· $h_{max}$

u² = 2 × 9.8 × 0.5 = 9.8

u = √9.8 ≈ 3.13

u = 3.13 m/s

Her initial jumping velocity ≈ 3.13 m/s

Escape velocity, $v_e = \sqrt{\dfrac{2 \cdot G \cdot M}{r} }$

Where;

M = The mass of the asteroid

G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

r = The radius of the asteroid

The average density of the Earth = 5515 kg/m³

The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³

The escape velocity, she has, $v_e$ ≈ 3.13 m/s is therefore;

$3.13 = \sqrt{\dfrac{2 \times 6.67408 \times 10^{-11} \times 5515 \times \frac{4}{3} \times \pi \times r^3}{r} } = r \times \sqrt{3.084 \times 10^{-6}}$

$r = \dfrac{3.13}{ \sqrt{3.084 \times 10^{-6}}} \approx 1782.45$

Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.

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3 years ago

List 5 possible effects of not adhering to standards of measurement

murzikaleks [220]

Answer:

factual evidence of customer-service levels.

better understanding of cross-functional performance.

enhanced alignment of operations with strategy.

evidence-based determination of process improvement priorities.

detection of performance trends.

better understanding of the capability range of a process.

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