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victus00 [196]
3 years ago
8

To determine the location of the center of mass (or center of gravity) of a car, the car is drivenover a scale on a horizontal f

loor. When the front wheels are over the scale, the weight recordedby the scale is 5800 N, and when the rear wheels are over the scale, the scale reads 6500 N. Thedistance between the front and rear wheels is measured to be 3.20 m. How far behind the frontwheels is the center of mass located?A) 1.59 m B) 1.50 m C) 1.72 m D) 1.69 m E) 0.845 m
Physics
1 answer:
just olya [345]3 years ago
4 0

Explanation:

It is given that,

When the front wheels are over the scale, the weight recorded by the scale is 5800 N, F₁ = 5800 N

When the rear wheels are over the scale, the scale reads 6500 N, F₂ = 6500

The distance between the front and rear wheels is measured to be 3.20 m, x₂ = 3.2 m

We need to find the location of center of mass behind the front wheels. Let the center of is located at a distance of x₁. Thus balancing the torques we get :

5800\times x_2=6500\times (3.2-x_2)

On solving the above equation we get, x₂ = 1.69 m

So, the center of mass is located at a distance of 1.69 meters behind the front wheels.

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Answer:

w = 1.976 rpm

Explanation:

For simulate the gravity we will use the centripetal aceleration a_c, so:

a_c = w^2r

where w is the angular aceleration and r the radius.

We know by the question that:

r = 60.5m

a_c = 2.6m/s2

So, Replacing the data, and solving for w, we get:

2.6m/s = w^2(60.5m)

W = 0.207 rad/s

Finally we change the angular velocity from rad/s to rpm as:

W = 0.207 rad/s = 0.207*60/(2\pi)= 1.976 rpm

3 0
4 years ago
The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential dif
Dovator [93]
The answer is B, because it will lose potential energy.
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3 years ago
A baseball thrown from the outfield to home plate does not have which of the following types of energy while it's in the air? A.
posledela
The ball does not have D. Radiant energy.
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block with of mass m is at rest on horizontal frictionless surface at time t=0. A force given by F=Bt+C is applied horizontally
Alexeev081 [22]

Answer:

v_{2} =\frac{1}{2}

Explanation:

From the second law of Newton movement laws, we have:

F=m*a, and we know that a is the acceleration, which definition is:

a=\frac{dv}{dt}, so:

F=m*\frac{dv}{dt}\\\frac{dv}{dt}=\frac{F}{m}=\frac{\frac{1}{2}(t+1)}{4}=\frac{t+1}{8}

The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):

dv=\frac{1}{8}(t+1)dt\\\int\limits^{v_{2}}_0 \, dv=\int\limits^{2}_{0} {\frac{1}{8}(t+1)} \, dt

v_{2}=\frac{1}{8}*(\frac{t^{2}}{2}+t) (This is the indefinite integral), the definite one is:

v_{2}=\frac{1}{8}*(2+2)=\frac{1}{2}

3 0
3 years ago
A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle
Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of <em>θ</em> = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height <em>y</em> at time <em>t</em> would be

<em>y</em> = (40 m/s) <em>t</em> - 1/2 <em>g t</em> ²

Set <em>y</em> = 160 m, and you'll find that there is no (real) solution for<em> t</em>, so the ball never attains the given maximum height.

From another perspective: recall that

<em>v </em>² - <em>v</em>₀² = 2<em>a </em>∆<em>y</em>

where

• <em>v</em>₀ = initial velocity

• <em>v</em> = final velocity

• <em>a</em> = acceleration

• ∆<em>y</em> = displacement

At its maximum height, the ball has zero vertical velocity, and ∆<em>y</em> = maximum height = 160 m. The ball is in free fall once it's launched, so <em>a</em> = -<em>g</em>.

So we have

0² - (40 m/s)² = -2<em>g </em>(160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0
3 years ago
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