What's the momentum of a 3.5-kg boulder rolling down hill at 5 m/s

Find the mass of a person walking west at a speed of 0.8 m/s with a momentum of 52.0 kg.m/s west.

KIM [24]

Answer:

mass of the person walking to west is 65 kg.

Given:

Momentum = 52 $\frac{kg m}{s}$

Speed = 0.8 $\frac{m}{s}$

To find:

Mass of the person = ?

Formula used:

Momentum is given by,

P = m × v

Where, P = momentum

m = mass

v = speed

Solution:

Momentum is given by,

P = m × v

Where, P = momentum

m = mass

v = speed

Mass = $\frac{P}{v}$

m = $\frac{52}{0.8}$

m = 65 kg

Thus, mass of the person walking to west is 65 kg.

5 0

1 year ago

Sea breezes that occur near the shore are attributed to a difference between land and water with respect to what property?

ddd [48]

Answer:

a. mass density

Explanation:

<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>

When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.

<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>

7 0

2 years ago

Is v2 = v1t+a dimensionally correct? Explain please!

Lady bird [3.3K]

You want v2 = v1 + at
v is measured in m/s, a in m/s2, and t in s.
the dimensions multiply like algebraic quantities.
so because v2 is measured in m/s, then (v1 + at) has to come out in m/s
the units for (v1 + at) are (m/s) + (m/s2)(s)
time "s" cancels out one acceleration "s", so it comes ut to (m/s) + (m/s), which = (m/s).
if you had (v1t + a), then you would have (m/s)(s) + (m/s2) which = (m) + (m/s2), which doesn't work.

4 0

1 year ago

If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he

shutvik [7]

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

$F = \frac{GMm}{R^2} = mg$

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

$(mg)'= \frac{GMm}{(R+h)^2}$

we need to find "h"

taking the ratio of two:

$\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km$

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

5 0

2 years ago

A 1000 kg roller coaster begins on a 10 m tall hill with an initial velocity of 6m/s and travels down before traveling up a seco

balu736 [363]

Answer:

10.6 meters.

Explanation:

We use the law of conservation of energy, which says that the total energy of the system must remain constant, namely:

$\frac{1}{2}mv_i^2+mgh_i-1700j=\frac{1}{2}mv_f^2+mgh_f$

In words this means that the initial kinetic energy of the roller coaster plus its gravitational potential energy minus the energy lost due to friction (1700j) must equal to the final kinetic energy at top of the second hill.

Now let us put in the numerical values in the above equation.

$m=100kg$