A body of mass 3.0kg moves with a velocity 10 m/s. calculate the momentum of the body

Cyclist always bends when moving the direction opposite to the wind. Give reasons

Nuetrik [128]

When he bends he kinda off his feet and light but if it’s not i’m so sorry this is just my thinking.

6 0

2 years ago

Sean pushes a box with a force of 100 N. Christian pushes an identical box with a force of

kap26 [50]

Answer:

Christian

Explanation:

Given that

Force applied by Sean ,F₁=100 N

Force applied by Christian ,F₂=200 N

Lets take mass of the box = m kg

As we know that from second law of Newton's

F= m a

F=Force,m=mass ,a=acceleration

$F_1=ma_1$

$100=m\times a_1$

$a_1=\dfrac{100}{m}\ m/s^2$

$F_2=ma_2$

$200=m\times a_2$

$a_2=\dfrac{200}{m}\ m/s^2$

From the above we can say that acceleration a₂ is greater than a₁ is for the same mass of the box.

Therefore Christian will acceleration more for the box.

5 0

3 years ago

Near the surface of the Earth, objects in free fall (but not terminal velocity) experience

Aleks [24]

B (9.81 m/s^2)
Speed no, because acceleration isn't 0
Velocity, pretty much same as speed
Distance no, because it's getting closer

5 0

2 years ago

Read 2 more answers

Define the focus of a concave lens

Sophie [7]

Answer:

Principal focus of concave lens - 

★ The point at which rays parallel to principal axis coming from infinity appear to converge after being refracted from concave lens is called the principal focus of concave lens.

________________________

• Additional information - 

★ Principal focus - A number of rays parallel to the principal axis after reflection from a concave mirror meet at a point on the principal axis or appear to come from a point after reflection from a convex mirror on the principal axis. This is called principal focus.

8 0

2 years ago

A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied

erastova [34]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

$\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\$

Calculating Shear Transverse:

$\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}$

$= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi$

$= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in$

$\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\$

$= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi$

8 0

2 years ago

A body of mass 3.0kg moves with a velocity 10 m/s. calculate the momentum of the body​

A body of mass 3.0kg moves with a velocity 10 m/s. calculate the momentum of the body