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3 years ago

7

Bumper car A (331 kg) moving east at 3.87 m/s hit stationary car B (208 kg). After, the car A moves .88 m/s east. What is the ve

locity of car B?

1 answer:

Gre4nikov [31]3 years ago

4 0

Answer:

Ub = 221.99m/s

Explanation:

(Ma x Ua)+(Mb x Ub) = (Ma + Mb)V

(331 x 3.87)+(208 x Ub) = (331 + 208)88

(1257.8) + (208 x Ub) = 47432

(208 x Ub) = 47432 - 1257.8

(208 x Ub) = 46174.2

Ub = 46174.2/208 = 221.99m/s

Ub = 221.99m/s

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Suppose a 20-foot ladder is leaning against a building, reaching to the bottom of a second-floor window 15 feet above the ground

Papessa [141]

Answer:

The answer is β=0,85 rads

Explanation:

As the ladder is leaning against the building, we can imagine there´s a triangle where 20ft is the hypotenuse and 15ft is the maximum vertical distance between the ladder and the ground, it means, the leg opposite to β which is the angle we need

Let β(betha) be the angle between the ladder and the ground

We also know that $sin(betha)=(leg opposite)/(hypotenuse)$

In this case we will need to find β, this way:

$betha=sin^-1((15ft/20ft))$

Then β=48,6°

We also have that 2πrads is equal to 360°, in this way we find how much β is in radians:

$betha=(48,6°)*(2pirads/360°)$

then we find β=0,85rads

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3 years ago

After robbing a bank, a criminal tries to escape from the police by driving at a constant speed of 55 m/s (about 125 mph). A pol

vfiekz [6]

Answer:

18.03 s

Explanation:

We have two different types of motions, the criminal moves with uniform motion while the police do it with uniformly accelerated motion. Therefore we will use the equations of these cases. We know that by the time the police reach the criminal they will have traveled the same distance.

$x=vt\\x=x_{0}+v_{0}t+\frac{a}{2}t^2$

The distance between the police and the criminal when the first one starts the persecution is 0, its initial speed is also zero. So:

$x=(55m/s)t\\x=\frac{6.1m/s^2}{2}t^2=(3.05m/s^2)t^2$

Equalizing these two equations and solving for t:

$(55m/s)t=(3.05m/s^2)t^2\\(3.05m/s^2)t^2-(55m/s)t=0\\t((3.05m/s^2)t-55m/s)=0\\t=0 \\(3.05m/s^2)t-55m/s=0\\t=\frac{55m/s}{3.05m/s^2}=18.03 s$

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3 years ago

A motorcycle accelerates uniformly from rest at 7.9\,\dfrac{\text{m}}{\text{s}^2}7.9 s 2 m 7, point, 9, space, start fraction,

8090 [49]

Answer:

t = 3.516 s

Explanation:

The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:

$v(t) = v_0 +at$

Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0

If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.

This can be done knowing that

1 km = 1000 m

1 h = 3600 s

Therefore

1 km/h = (1000/3600) m/s = 0.2777... m/s

100 km/h = 27.777... m/s

Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:

27.777 m/s = 7.9 m/s^2 t

Solving for t

t = (27.7777 / 7.9) s = 3.516 s

6 0

3 years ago

The kinetic energy of a 1000-kg roller coaster car that is moving with a speed of 20.0 m/s.

eimsori [14]

E=(mV^2)/2
m=1000kg, V=20m/s
then, E=(1000kg*(20m/s)^2)/2
E=(1000*400)/2 J = 200000J

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3 years ago

A pitcher throws a baseball that reaches the catcher in 0.75 s. The ball curves because it is spinning at an average angular vel

7nadin3 [17]

The change in angular displacement as a function of time is the definition given for angular velocity, this is mathematically described as

$\omega = \frac{\theta}{t}$

Here,

$\theta$ = Angular displacement

t = time

The angular velocity is given as

$\omega = 230rev/min$

PART A) The angular velocity in SI Units will be,

$\omega = 230rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega = \frac{23}{3}\pi rad/s \approx 24.08rad/s$

PART B) From our first equation we can rearrange to find the angular displacement then

$\theta = \omega t$

Replacing,

$\theta = (24.08)(0.75)$

$\theta = 18.06 rad$

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3 years ago