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andrezito [222]
2 years ago
10

When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, th

e magnitude of the normal force of the floor on the crate isA. 25N
B. 68N
C. 180N
D. 250N
E. 310N

Physics
1 answer:
Fofino [41]2 years ago
3 0

Answer:

Option E is correct 310N

Explanation:

Given that the force used to push the crate is F = 200N

The force directed 20° below the horizontal

Mass of crate is m = 25kg

Weight of the crate can be determine using

W = mg

g is gravitational constant =9.8m/s²

W = 25×9.8

W = 245 N

Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

ΣFy = m•ay

ay = 0, since the body is not moving in the vertical or y direction

N—W—F•Sin20 = 0

N = W+F•Sin20

N = 245+ 200Sin20

N = 245 + 68.4

N = 313.4 N

The normal force is approximately 310 N to the nearest ten

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A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"
Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

4 0
2 years ago
Which law is associated with inertia
N76 [4]

Answer:

The focus of Lesson 1 is Newton's first law of motion - sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

4 0
3 years ago
a 252,000j of heat energy is required to heat 2kg water. if (CH2O=42000j/kgk.what is change in temperature ?​
navik [9.2K]

Answer:

theres an decrease in temperature because 252,000 is more than 42,000. so its colder and not as hot as 252,000.

Explanation:

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Galina-37 [17]

Answer:

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Explanation:

7 0
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When a compound, such as salt, dissolves in water, it's said to be:
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Answer:D:soluble

Explanation:

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5 0
3 years ago
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