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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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Answer:

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density = 10^{18} kg/m³

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1,520.00 calories

Explanation:

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Given that ĉ = Q / (m.ΔT),

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A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The t
azamat

Answer:

(a) x'(t)= 142

(b) 142

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(e) 0.426 s

(f) 0.061 rad

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Velocity is a time-derivative of position.

(a) x(t) = 142t

x'(t)= 142

(b) Since x'(t)= 142 is independent of t, it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

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In radians, this is

\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}

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