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kiruha [24]
3 years ago
9

A forward horizontal force of 50 N is applied to crate a second horizontal force of 180 N is applied to crate in the opposite di

rection determine the magnitude and direction of the resultant force acting on the crate
​
Physics
1 answer:
e-lub [12.9K]3 years ago
8 0

Answer:

130n on the 2nd horizontal

Explanation:

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A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio
Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be 8.21\times 10^3~N

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

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A device is rated at 1.3kW when connected to a 120 V source. The equivalent resistance of this device in ohm is:
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D.

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The graph represents the reaction 2H2 + 02 32H20 as it reaches
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C and D

Explanation:

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3 years ago
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How is acceleration calculated?
Xelga [282]

Answer:

Acceleration equals the change in velocity divided by time.

Explanation:

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4 years ago
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Answer:

Explanation:

Given

Inclination \theta =32.9^{\circ}

Distance of landing point R=7.78\ m

Considering athlete to be an Projectile

range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

where u=launch velocity

7.78=\frac{u^2\sin (2\times 32.9)}{9.8}

u^2=83.58

u=\sqrt{83.58}

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R'=9.07\ m

 

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