All categories

3 years ago

Explain what will happen if an object has high momentum or low momentum and a force is applied?

1 answer:

5 0

-- If the force is applied in the same direction as the object is moving, then the object's momentum in that direction will increase.

-- If the force is applied in the direction OPPOSITE to the way the object is moving, then the object's momentum will decrease.

-- In either case, the CHANGE in the object's momentum will be

(strength of the force) x (length of time the force is applied) .

This quantity is also called "impulse".

You might be interested in

As a roller coaster car crosses the top of a 50-m-diameter loop-the-loop, its apparent weight is the same as its true weight.

scZoUnD [109]

Answer:22.36 m/s

Explanation:

Given

the diameter of loop d=50 m

the radius of loop r=25 m

At the top position, we can write,

weight and Normal reaction combination will provide the centripetal force i.e.

$R+W=\frac{mv^2}{r}$

$R=W\quad \quad [\text{apparent weight =Actual weight}]$

$2W=2mg=\frac{mv^2}{r}$

$v=\sqrt{2gr}$

$v=\sqrt{2\times 10\times 25}$

$v=22.36\ m/s$

3 0

3 years ago

Can anyone explain and get the answer pls?

Veronika [31]

FL₁ =Fl₂

80.x₁ = 30.7

x₁=2.625 m

6 0

2 years ago

A rock is thrown straight up with an initial velocity of 15.0 m/s. Ignore energy lost to air friction. How high will the rock ri

Furkat [3]

$h=v_0t- \frac{1}{2} gt^2\\h=15t-\frac{1}{2}(9.8) t^{2}\\ \\ 
v_f=v_0-gt\;\;\;\Rightarrow t= \frac{v_f-v_0}{-g} = \frac{0-15}{-9.8} \approx1.53 \\ \\ 
\Rightarrow h=15(1.53)-4.9(1.53)^2\;\;\;\Rightarrow h=11.48\,m$

5 0

3 years ago

A body of mass 70kg climbed a hill 1000m high. calculate the maximum energy gained by the body.

stellarik [79]

Potential energy = mgh

So, energy gained

= mgh

= 70kg × (9.8m/s²) × 1000m

= 686000 kgm²/s²

= 686000 J

3 0

3 years ago

A quantity of N2 occupies a volume of 1.4 L at 290 K and 1.0 atm. The gas expands to a volume of 3.3 L as the result of a change

lions [1.4K]

Answer:

$\rho = 0.50 g/L$

Explanation:

As we know that

PV = nRT

here we have

$P = 1.0 atm$

$P = 1.013 \times 10^5 Pa$

so we have

$V = 1.4 \times 10^{-3} m^3$

T = 290 K

now we have

$(1.013 \times 10^5)(1.4 \times 10^{-3}) = n(8.31)(290)$

$n = 0.06$

now the mass of gas is given as

$m = n M$

$m = (0.06)(28)$

$m = 1.65 g$

now density of gas when its volume is increased to 3.3 L

so we will have

$\rho = \frac{m}{V}$

$\rho = \frac{1.65 g}{3.3 L}$

$\rho = 0.50 g/L$

5 0

3 years ago