All categories

3 years ago

Is momentum proportional or inversely proportional to mass?

1 answer:

3 0

Answer:In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object. where m is the mass and v is the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.

Explanation:

You might be interested in

16x^2y^2-25a^2b^2<br>factorize the expression

SIZIF [17.4K]

Answer:

(4xy+5ab)(4xy-5ab)

Explanation:

16 $x^{2}$ $y^{2}$ -25 $a^{2}$ $b^{2}$

4^2 is 16 and 5^2 is 25,

Also, (x-a)(x+a) = x^2-a^2

So, this factorized is:

(4xy+5ab)(4xy-5ab)

Hope this helps!

8 0

3 years ago

A 21.0 kg shopping cart is moving with a velocity of 6.0 m/s. It strikes a 11.0 kg box that is initially at rest. They stick tog

DedPeter [7]

Answer:

a) 126 kgm/s

b) 0 kgm/s

c) 3.9 m/s

Explanation:

To solve this question, we will use the law of conservation of momentum.

Momentum before collision = momentum after collision

m1v1 + m2v2 = (m1 + m2)v, where

m1 = mass of the shopping cart, 21 kg

m2 = mass of the box, 11 kg

v1 = initial velocity of the shopping cart, 6 m/s

v2 = initial velocity of the box, 0 m/s

v = final velocity of the box+cart

Momentum of the shopping cart before collision = P

P = mv

P = 21 * 6

P = 126 kgm/s = c

Momentum of the box before collision

Like in question a above, the momentum of the box is P

P = mv

P = 11 * 0

P = 0 kgm/s = b

Velocity of the combined shopping cart wreckage after collision is

m1v1 + m2v2 = (m1 + m2)v

(21 * 6) + (11 * 0) = (21 + 11)v

126 + 0 = 32v

32v = 126

v = 126/32

v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.

Please vote brainliest

8 0

3 years ago

Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.

Naily [24]

(a) 0.249 (24.9 %)

The maximum efficiency of a heat engine is given by

$\eta = 1-\frac{T_C}{T_H}$

where

Tc is the low-temperature reservoir

Th is the high-temperature reservoir

For the engine in this problem,

$T_C = 270^{\circ}C+273=543 K$

$T_H = 450^{\circ}C+273=723 K$

Therefore the maximum efficiency is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249$

(b-c) 0.221 (22.1 %)

The second steam engine operates using the exhaust of the first. So we have:

$T_H = 270^{\circ}C+273=543 K$ is the high-temperature reservoir

$T_C = 150^{\circ}C+273=423 K$ is the low-temperature reservoir

If we apply again the formula of the efficiency

$\eta = 1-\frac{T_C}{T_H}$

The maximum efficiency of the second engine is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221$

8 0

3 years ago

Suppose the displacement of an object is related to time according to the expression x=By*2, what are the dimensions of B

laiz [17]

Answer:

Explanation:

AP3X

5 0

3 years ago

Which best describes the motion of air particles when a transverse wave passes through them?

malfutka [58]

Answer is c that is he particles move perpendicular to the direction of the wave.

8 0

4 years ago