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9966 [12]
3 years ago
14

What is warmer during the night: land or sea/ocean?

Physics
2 answers:
kupik [55]3 years ago
5 0
I do believe the answer is land because the ocean/sea gets cold when its night. So the answer is land
Phantasy [73]3 years ago
4 0
I am pretty sure that it's land
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A stone is launched from the ground, at a 70° angle, with an initial velocity of 120 m/s.
zavuch27 [327]
<span>A) x = 41t
    The classic equation for distance is velocity multiplied by time. And unfortunately, all of your available options have the form of that equation. In fact, the only difference between any of the equations is what looks to be velocity. And in order to solve the problem initially, you need to divide the velocity vector into a vertical velocity vector and a horizontal velocity vector. And the horizontal velocity vector is simply the cosine of the angle multiplied by the total velocity. So H = 120*cos(70) = 120*0.34202 = 41.04242 So the horizontal velocity is about 41 m/s. Looking at the available options, only "A" even comes close.</span>
3 0
3 years ago
Read 2 more answers
Ben hit a 0.25kg nail into a wood with a 5.2kg hammer. If the hammer moves witht the speed of 52 m/s and two fifth of its kineti
Art [367]

Answer:

Explanation:

Mass of nails is 0.25kg

Mass of hammer 5.2kg

Speed of hammer is =52m/s

Then, Ben kinetic energy is given as

K.E= ½mv²

K.E= ½×5.2×52²

K.E= 7030.4J

Given that, two-fifth of kinetic energy is converted to internal energy

Internal energy (I.E) = 2/5 × K.E

Internal energy (I.E) = 2/5 × 7030.4

I.E=2812.16J.

Energy increase is total Kinetic energy - the internal energy

∆Et= K.E-I.E

∆Et= 7030.4 - 2812.16

∆Et= 4218.24J

7 0
3 years ago
the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force
-Dominant- [34]

Answer:

PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

F_{net} = 51 N

now we also know that friction force due to ground is given as

F_f = 24 N

now we have

F_{net} = F_{ext} - F_f

51 = F_{ext} - 24

F_{ext} = 75 N

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

a = \frac{F_f}{m}

a = \frac{24}{24} = 1 m/s^2

now we can find the distance by kinematics

v_f^2 - v_i^2 = 2 a d

0 - 1.5^2 = 2(-1)d

d = 1.125 m

so the distance moved will be 1.125 m

6 0
3 years ago
2 Which is true of a parallel circuit?
mars1129 [50]

Answer:

fxb

c

Explanation:bfffffffff

8 0
3 years ago
31. Draw a free body diagram for a 15.5N box that is being pushed to the right with a 18. N force while experiencing 4.30 N of r
posledela

Answer:

See answers below

Explanation:

a.

F = mg,

15.5 N = m(9.8 m/s²)

m = 1.58 kg

b.

Fnet = Applied force - resistance,

Fnet = 18 N - 4.30 N,

Fnet = 13.70 N

Fnet = ma

13.70 N = (1.58 kg)a

a = 8.67 m/s²

For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.

7 0
2 years ago
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