Does exists the friction force in space?If yes,tell an full example

A 100 kg father and his 50 kg daughter stand on ice skates on a frozen pond. The daughter pushes on her father with a force of 5

Lana71 [14]

Answer:

Both feel same magnitude force.

Explanation:

Given that

Mass of father = 100 kg

Mass of daughter = 50 kg

From the third law of Newtons :

Every action have it reaction in opposite direction.It means that if any object apply a force on the other object then first object also feel force of same magnitude but in opposite direction.

It means that both father and her daughter will feel same magnitude of same but in opposite direction.

7 0

2 years ago

Read 2 more answers

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

2 years ago

One of your classmates, in a fit of unrestrained ego, jumps onto a lab table:

Anettt [7]

The equilibrium condition allows finding the results for the forces of the system are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition

∑ F = 0

Where the bold letters indicate vectors, F is the external forces

a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.

b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force

N = N₁ + N₂ + N₃ + N₄₄

Let's apply the equilibrium condition

N - $W_m -w_{table}$ = 0

N = $W_m +w_{table}$

N = $M_m g + w_{table}$

They indicate the pose of the boy is 65 kg, for the weight of the table of a laboratory table is approximately 100 N

N = 65 9.8 + 100

N = 737 N

c) To calculate the mass of the table we use the relation

W = $m_{table}$ g

$m_{table} = \frac{w_{table}}{g}$

$m_{tabble}= \frac{100}{9.8}$

$m_{table}$ e = 10.2 kg

In conclusion using the equilibrium condition we can find the results for the forces are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Learn more here: brainly.com/question/19860811

7 0

2 years ago

How long will it take to travel 200 m traveling at 10 m/s? Follow example below.

polet [3.4K]

Answer:

$variables - d = 200m \: v = 10 m {s}^{ - 1} \\ equation \: - v = \frac{d}{t} \\ 10 = \frac{200}{t} \\ cross \: multiply \\ 10t = 200 \\ \frac{10t}{10} = \frac{200}{10} \\ t = 20s$