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bagirrra123 [75]
3 years ago
13

A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of

0.5 m/s2 across a horizontal floor. What is the value of the normal force on the box? 200 N 243 N 156 N 225 N

Physics
1 answer:
PtichkaEL [24]3 years ago
8 0

Answer:

225 N

Explanation:

"Below the horizontal" means he's pushing down at an angle.

Draw a free body diagram of the box.  There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.

Sum of forces in the y direction:

∑F = ma

N − mg − F sin θ = 0

N = F sin θ + mg

Plug in values:

N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)

N = 225 N

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The cheetah, the fastest of land animals, can run a distance of 274 m in 8.65 s at its top speed. What is the cheetahs tips spee
ycow [4]

Answer:

Top speed is 31.68 m/s

Explanation:

Speed of a body is the distance traveled by the body in unit time.

Speed is generally expressed in meter per second or kilometer per hour.

Here, the cheetah travels a distance of 274 m in 8.65 s.

Therefore, speed of cheetah is the distance traveled by it in one second.

For 8.65 s, distance traveled is 274 m

For 1 s, distance traveled will be \frac{274}{8.65} meters per second. This value is called speed.

So, speed is given as the ratio of the distance traveled to that of the time taken.

∴ Top speed of cheetah is given as:

v_{max}=\frac{274}{8.65}=31.68 \textrm{ m/s}

Here, v_{max} is the top speed of cheetah.

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4 years ago
A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the
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Answer:

Explanation:

Check attachment for solution

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3 years ago
If Q = 16 nC, a = 3.0 m, and b = 4.0 m, what is the magnitude of the electric field at point P?
Lynna [10]
In BPC

tan\theta =a/b = 3/4

\theta = tan^-1(0.75)

\theta = 36.87 deg

BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m

Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C

Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C

Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C

Net electric field along X-direction is given as

Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C

Net electric field along X-direction is given as

Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C

Net electric field is given as

E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C
8 0
3 years ago
What are the harmful effects of drugs in the body?​
snow_lady [41]

Explanation:

causes hallucinations

causes lung cancer eg cigarettes

causes bad breath

damages the nervous system eg cocaine

6 0
3 years ago
Read 2 more answers
Wagon wheel. While working on your latest novel about settlers crossing the Great Plains in a wagon train, you get into an argum
OverLord2011 [107]

Answer:

I = 16.7kgm²

Explanation:

Since, Torque is given by,

\tau = F*r = I*\alpha

here, I = Moment of inertia = ??

\alpha = angular acceleration of wheel = a/r

F = tangential tension acting on the wheel = T

a = acceleration of bag of sand = 2.95 m/s^2

r = radius of wheel = d/2 = 120/2 = 60 cm = 0.60 m

from force balance on sand bag,

mg - T = m*a

T = m*(g-a)

m = mass of sand bag = 20 kg

So, I = T*r/\alpha = m*(g-a)*r/(a/r)

Using known values:

I = 20*(9.81 - 2.95)*0.60/(2.95/0.60) = 16.74

I = 16.7 kgm² = Moment of inertia of wheel experimentally

also, Moment of inertia of wheel theoretically(I') = M*r²

given, M = mass of wheel = 70 kg

I' = 70*0.60²= 25.2 kgm² = Moment of inertia of wheel theoretically

7 0
3 years ago
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