What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?

A cow has eaten 1500KJ of stored chemical energy in the form of food. 945KJ is exerted as waste products. 495kJ is used for resp

n200080 [17]

We need Net Energy

$\\ \rm\Rrightarrow E_{net}=1500-(945+495)$

$\\ \rm\Rrightarrow E_{net}=1500-1440$

$\\ \rm\Rrightarrow E_{net}=60kJ$

7 0

2 years ago

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Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.

GenaCL600 [577]

So, the force of gravity that the asteroid and the planet have on each other approximately $\boxed{\sf{2.9 \times 10^{17} \: N}}$

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

$\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}$

With the following condition :

F = gravitational force (N)
G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_1}$ = mass of the first object (kg)
$\sf{m_2}$ = mass of the second object (kg)
r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_X}$ = mass of the planet X = $\sf{1.55 \times 10^{22}}$ kg.
$\sf{m_Y}$ = mass of the planet Y = $\sf{3.95 \times 10^{28}}$ kg.
r = distance between two objects = $\sf{3.75 \times 10^{11}}$ m.

What was asked :

F = gravitational force = ... N

Step by step :

$\sf{F = G \times \frac{m_X \times m_Y}{r^2}}$

$\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}$

$\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}$

$\sf{F \approx 2.9 \times 10^{39 - 22}}$

$\sf{F \approx 2.9 \times 10^{17} \: N}$

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

$\boxed{\sf{2.9 \times 10^{17} \: N}}$

Gravity is a thing has depends on ... brainly.com/question/26485200

8 0

2 years ago

Two large aluminum plates are separated by a distance of 2.0 cm and are held at a potential difference of 170 V. An electron ent

notsponge [240]

Answer:

Explanation:

Electric field between plates

V / d

= 170 / ( 2 x 10⁻² )

= 8500 N/C

Force on electron in this field

= 8500 x 1.6 x 10⁻¹⁹

= 13600 x 10⁻¹⁹ N

Acceleration

= 13600 x 10⁻¹⁹ / 9.1 x 10⁻³¹

a = 1494.5 x 10¹² m /s²

s = .1 x 10⁻² m

v² = u² + 2as

= (2.9x 10⁵)²+ 2 x 1494.5 x 10¹² x .1 x 10⁻²

= 8.41 x 10¹⁰ + 299 x 10¹⁰

= (8.41 + 299 ) x 10¹⁰

v = 17.53 x 10⁵ m /s

8 0

3 years ago

Which requires more work, lifting a 10.0kg load a vertical distance of 2m or lifting a 5.0kg load a distance of 4m?

Minchanka [31]

For each load, Work = (mass) x (gravity) x (distance .

Bigger load: Work = (10 kg) x (9.8 m/s²) x (2 m) = 196 joules .

Smaller load: Work = (5 kg) x (9.8 m/s²) x (4 m) = 196 joules.

The work required is equal in both cases.

The mass ratio of 2:1 is exactly balanced by
the height ratio of 1:2 .

6 0

3 years ago

What is the magnitude of the force that is exerted on a 20 kg mass to give it an acceleration of 10.0

ANEK [815]

Answer:Mass of the body = 20 kg.

Final Velocity = 5.8 m/s.

Initial velocity = 0

Time = 3 seconds.

Using the Formula,

Acceleration = (v - u)/ t

= (5.8 - 0)/ 3

= 1.6 m /s².

Now, Using the Formula,

Force = mass × acceleration

= 20 × 1.6

=

Explanation: I REALLY HOPE THIS HELPS I'M KINDA NEW AT THIS :] :]

8 0

2 years ago