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lesantik [10]
3 years ago
10

What is the momentum in Kg m/s of a 10 kg truck travelling at A) 5 m/s B) 20 cm/s C) 36 km/h?​

Physics
1 answer:
san4es73 [151]3 years ago
7 0

Answer:

A)   50 kg m/s

B)  2 kg m/s

C)  100 kg m/s

Explanation:

Recall that momentum is a physical quantity defined by the product of the object's mass times its velocity. Therefore, in our case the truck's momentum (p) is:

A)  p = m * v = 10 kg * 5 m/s = 50 kg m/s

B) p = m * v = 10 kg * 0.20 m/s = 2 kg m/s (notice that the velocity in cm/s was converted into m/s before multiplying)

C) the conversion from km/h to m/s is a little more involved, so we taclke it first:

since 1 km = 1000 m  and 1 hour = 3600 seconds, we convert 36 km/h by replacing each unit by its m and s expression as shown below:

36 \frac{km}{h} =36\,\frac{1000\,m}{3600\,\,s}=10\,\frac{m}{s}

Then, the momentum is:  p = m * v = 10 kg * 10 m/s = 100 kg m/s

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The mass of the child in this problem is 24 Kg, the unit for mass is in Kilogram.
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Nesterboy [21]
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3 years ago
Using the strap at an angle of 31.0° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15.0
lora16 [44]

Answer:

<h2>154.73N</h2>

Explanation:

The question is incomplete. Here is the complete question.

Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.

Check the diagram related to the question in the attachment below for better understanding.

The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.

The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).

Ty = 15sin31°

Ty = 7.73N

W = mass * acceleration due to gravity

W = 15.0*9.8

W = 147N

The normal force is therefore expressed as;

N = Ty + W

N = 7.73 + 147

N = 154.73N

3 0
3 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
iren [92.7K]

Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

E=Pt=(9.36)(350)=3276 J

Finally, the change in temperature of an object is related to the energy supplied by

E=C\Delta T

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

Solving for C, we find:

C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

5 0
2 years ago
A radioactive material has a count rate of 400 per minute. It has a half life of 40 years. How long will it take to decay to a r
cestrela7 [59]

Answer:

160 years.

Explanation:

From the question given above, the following data were obtained:

Initial count rate (Cᵢ) = 400 count/min

Half-life (t½) = 40 years

Final count rate (Cբ) = 25 count/min

Time (t) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Initial count rate (Cᵢ) = 400 count/min

Final count rate (Cբ) = 25 count/min

Number of half-lives (n) =?

Cբ = 1/2ⁿ × Cᵢ

25 = 1/2ⁿ × 400

Cross multiply

25 × 2ⁿ = 400

Divide both side by 25

2ⁿ = 400/25

2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time taken for the radioactive material to decay to the rate of 25 counts per minute. This can be obtained as follow:

Half-life (t½) = 40 years

Number of half-lives (n) = 4

Time (t) =?

n = t / t½

4 = t / 40

Cross multiply

t = 4 × 40

t = 160 years.

Thus, it will take 160 years for the radioactive material to decay to the rate of 25 counts per minute.

7 0
2 years ago
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