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3 years ago

Which is the momentum of a 100 kilogram boulder rolling south down a hill at a velocity of 5 meters per second

1 answer:

6 0

P = m * v
P = 100 * 5 Kg-m/s towards south
P = 500 Kg-m/s towards south

In short, your answer would be 500 Kg-m/s towards south

Hope this helps!

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Neither _____ nor _____ can be created or destroyed, but they can be changed from one to the other

Viktor [21]

Matter nor energy

hope this helps

3 0

4 years ago

Read 2 more answers

20 POINTS TRUE OR FALSE:

romanna [79]

False because the first law of motion is the crash I think I’m in middle school

8 0

3 years ago

An electric heater draws a steady 15.0A on a 120-V

morpeh [17]

Answer:

1. 1800 W

2. $ 17.3

Explanation:

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

Current (I) = 15 A

Voltage (V) = 120 V

Time (t) = 20 h per day

Duration = 31 days

Cost = 15.5 cents per kWh

1. Determination of the power.

Current (I) = 15 A

Voltage (V) = 120 V

Power (P) =?

P = IV

P = 15 × 120

P = 1800 W

Thus, 1800 W of power is required.

2. Determination of the cost per month (31 days).

We'll begin by converting 1800 W to KW.

1000 W = 1 KW

Therefore,

1800 W = 1800 W × 1 KW / 1000 W

1800 W = 1.8 KW

Next, we shall determine the energy consumption for 31 days. This can be obtained as follow:

Power (P) = 1.8 KW

Time (t) = 2 h per day

Time (t) for 31 days = 2 × 31 = 62 h

Energy (E) =?

E = Pt

E = 1.8 × 62

E = 111.6 KWh

Finally, we shall determine the cost of consumption. This can be obtained as follow:

1 KWh = 15.5 cents

Therefore,

111.6 KWh = 111.6 KWh × 15.5 cents / 1 KWh

111.6 KWh = 1729.8 cents

Converting 1729.8 cents to dollar, we have:

100 cents = $ 1

Therefore,

1729.8 cents = 1729.8 cents × $ 1 / 100 cents

1729.8 cents = $ 17.3

Thus, it will cost $ 17.3 per month to run the electric heater.

4 0

3 years ago

A block with mass M is placed on an inclined plane with slope angle q and is connected to a second hanging block with mass m by

tensa zangetsu [6.8K]

Answer:

The mass of the block m is:

$m=M(sin(\theta)+\mu_{s}cos(\theta))$

Explanation:

Let's analyze the block by parts

For the block M

$T-W_{x}-f_{f}=0$ (1)

Where:

T is the tension
W(x) is the component of the weight in the x-direction
F(f) is the friction force

$T-Mgsin(\theta)-\mu_{s}N=0$

$T-Mgsin(\theta)-\mu_{s}Mgcos(\theta)=0$

For the block m

$T-W=0$

$T=mg$ (2)

Now, let's combines equation (1) and (2):

$mg-Mgsin(\theta)-\mu_{s}Mgcos(\theta)=0$

Finally, let's solve it for block m.

$mg-Mg(sin(\theta)+\mu_{s}cos(\theta))=0$

$m=M(sin(\theta)+\mu_{s}cos(\theta))$

I hope it helps you!

7 0

3 years ago

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeof

BlackZzzverrR [31]

<h2>Answer: 26,8 s</h2>

Explanation:

If we are talking about an acceleration at a constant rate , we are dealing with constant acceleration, hence we can use the following equations:

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (1)

$V_{f}=V_{o}+at$ (2)

Where:

$V_{f}$ is the final velocity of the plane (the takeoff velocity in this case)

$V_{o}=0$ is the initial velocity of the plane (we know it is zero because it starts from rest)

$a=5m/s^{2}$ is the constant acceleration of the plane to reach the takeoff velocity

$d=1800m$ is the distance of the runway

$t$ is the time

Knowing this, let's begin with (1):

${V_{f}}^{2}=0+2(5m/s^{2})(1800m)$ (3)

${V_{f}}^{2}=18000m^{2}/s^{2}$ (4)

$V_{f}=134.164 m/s$ (5)

Substituting (5) in (2):

$134.164 m/s=0+(5m/s^{2})t$ (6)

Finding $t$ :

$t=26.8 s$ This is the time needed to take off

6 0

3 years ago