How does mass affect the gravitational pull of the Sun, Earth, and the Moon?

Un'automobile di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

sesenic [268]

Answer:

e di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiute di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiut

Explanation:

e di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiute di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiut

3 0

2 years ago

If an object weighs 80 Newtons, what force must be applied to the rope to raise the object when the mechanical advantage is four

AnnyKZ [126]

80 -4f(45.56 - 56) solve and get answer

5 0

2 years ago

The bottom of cooking utensils are often kept black. Give reason

stepan [7]

Answer:

absorb more heat and get heated quickly

Explanation:

7 0

3 years ago

Indiana jones (83.5 kg) is running 3.75 m/s when he jumps in a stationary 312 kg mine cart. what is their joint velocity afterwa

Lubov Fominskaja [6]

Answer:

.7917 m/s

Explanation:

This is a conservation of momentum question. You have an object initially at rest (cart) so that object is initially at 0 momentum. Indiana Jones is 83.5 kg and running 3.75 m/s so he starts with a momentum of 313.125 kg * m/s because momentum is equal to mass * velocity. Once the person jumps in the cart, the cart and the person can be considered one object and by conservation of momentum, the momentum of the Indiana-cart system is equal to 313.125 kg * m/s. By that, we can set that momentum equal to the combined mass * joint velocity. So 313.125 = (83.5kg + 312kg) * joint velocity. Then just solve for the velocity. The answer should be smaller than the intial velocity of the person of 3.75 m/s because the mine cart is HUGE at 312kg.

3 0

3 years ago

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

$F = \frac{k|q_1||q_2|}{r^2}$

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

4 0

3 years ago