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Harlamova29_29 [7]
3 years ago
9

What is the power of a snowplow moving a snow pile 18 meters with 800 N of force in 15 seconds?(Hint; two part question )

Physics
1 answer:
VMariaS [17]3 years ago
7 0

Answer:

960 W

Explanation:

F = 800 N

x = 18 m

t = 15 s

Work = Fx= (800 N)(18 m) = 14,400 J

Power = Work / time = (14,400 J) / (15 s) = 960 W

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An airplane has a takeoff speed of 62 m/s. assuming a constant acceleration of 1.7 m/s2, what is the minimum length of runway it
UkoKoshka [18]

Data:

u=0 m/s is the initial velocity of the plane

v=62 m/s is the final velocity of the plane (at which the plane takes off)

a=1.7 m/s^2 is the acceleration of the plane

To find the minimum distance S the plane needs to take off, we can use the following equation:

2aS=v^2 -u^2

Re-arranging it and substituting the numbers, we find

S=\frac{v^2-u^2}{2a}=\frac{(62 m/s)^2-0}{2(1.7 m/s^2)}=1131 m

3 0
3 years ago
How much force is needed to stop a 90-kg soccer player if he decelerates at 15 m/s²? A. 6 N B. 15 N C. 135 N D. 1350 N
Mariana [72]
The answer is: D. 1350 N

5 0
3 years ago
Read 2 more answers
Locate the complete verbal phrase and identify its type.
Ksju [112]
<span>won
adjective

Verb phrases are verbs that may function as a predicate, adjective, or adverb. </span>

(a) "That he said" is an adjective modifying "word". However, this contains the s ubject"he" and the verb "said". It is a clause and NOT a phrase. Phrases can only have either a verb or a noun.
<span>(b) There's only one verb "was" but it does not come with a complement, object, modifier, or other verb. Hence, it's NOT a verb phrase. </span>
<span>(c) "Shall be" consists of the modal shall and the be-verb be. This is a perfect example of a verb phrase that functions as a VERB PHRASE. </span>
<span>(d) "Roared" and "charged" are two verbs referring to different subjects. They do not come with a complement, object, modifier, or another verb. Hence, they're NOT a verb phrase. "As the bull charged" is a clause and not a phrase.</span>

7 0
3 years ago
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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0
3 years ago
A 50-gram sample of water is initially at a temperature of 22 °C. The sample is heated until the temperature is 32 °C The specif
forsale [732]

Answer:

500cal

Explanation:

Given parameters:

Mass of water  = 50g

Initial temperature  = 22°C

Final temperature  = 32°C

Specific heat of water  = 1cal/g

Unknown:

Amount of heat absorbed by the water in calories  = ?

Solution:

To solve this problem, we use the expression below:

       H  = m c Ф

H is the amount of heat absorbed

m is the mass

c is the specific heat capacity

Ф is the temperature change

       H  = 50 x 1 x (32  - 22)  = 500cal

5 0
2 years ago
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