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klemol [59]
3 years ago
14

A locomotive has a mass of 200,000kg. It is moving at 4.5m/s. Find its momentum .

Physics
2 answers:
anastassius [24]3 years ago
4 0
Momentum (p) = mass × velocity

P= 200,000×4.5

P= 900,000 .... answer !!
butalik [34]3 years ago
3 0
I agree, 900000 yay yay
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A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds
Inessa05 [86]

Answer:

A body travels 10 meters during the first 5 seconds of its travel,and a total of 30 meters over the first 10 seconds of its travel

20miles / 5sec = 4miles /sec would be the average speed for the last 20 m

Explanation:

The answer is 4 m/s.

In the first 5 seconds, a body travelled 10 meters. In the first 10 seconds of the travel, the body travelled a total of 30 meters, which means that in the last 5 seconds, it travelled 20 meters (30m + 10m).

The relation of speed (v), distance (d), and time (t) can be expressed as:

v = d/t

We need to calculate the speed of the second 5 seconds of the travel:

d = 20 m (total 30 meters - first 10 meters)

t = 5 s (time from t = 5 seconds to t = 10 seconds)

Thus:

v = 20m / 5s = 4 m/s

PLEASE GIVE BRAINIEST!! HOPE THIS HELPS

5 0
3 years ago
Suppose a car travels 106 km at a speed of 28 m/s and uses 1.9 gals of gasoline in the process. Only 30% of the gasoline goes in
USPshnik [31]

Answer:

a) The magnitude of the force is 968 N

b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N

Explanation:

<em>NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).</em>

Information given:

d = 106 km = 106,000 m

v1 = 28 m/s

G = 1.9 gal

η = 0.3

Eff = 1.2 x 10^8 J/gal

a) We can express the energy used as the work done. This work has the following expression:

W=F\cdot d

Then, we can derive the magnitude of the force as:

F=\frac{W}{d}=\frac{\eta\cdot (G\cdot Eff)}{d}=\frac{0.3*1.9*(1.8*10^8)}{106*10^3} =968\,N

b) We will calculate the force for a speed of 30 m/s.

If the force is proportional to the speed, we have:

F_2=F_1(\frac{v_2}{v_1} )=968(\frac{30}{28} )=968*1.0714=1,037\,N

6 0
2 years ago
Nora walks down a street and sees a ball dropped from a building
Grace [21]
It is gravity¿ what is the question?
5 0
3 years ago
During chemistry class, Carl performed several lab tests on two white solids. The results of three tests are seen in the data ta
Nastasia [14]

Answer: it’s c) ionic

Explanation:

6 0
3 years ago
Read 2 more answers
A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber
Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}

Now calculate the bouyant force acting on the lead

F_L = Vpg

F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N

This force will act in upward direction

Gravitational force on the lead due to its mass  will act in downward direction

Hence the difference of this two force

T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N

If V is the volume submerged in the water then bouyant force on the bobber is

F_B=V'pg

Equate bouyant force with the tension and gravitational force

F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}

Now Total volume of bobble is

\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\

=\large\boxed{4.52 \%}

7 0
3 years ago
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