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2 years ago

A car of mass 1200 kg, moving with a speed of 72 mph on a highway,

Physics

1 answer:

Keith_Richards [23]2 years ago

8 0

momentum(p) = mass(m) x velocity(v)

KE = kinetic energy = 1/2 mv²

the ratio of the momentum of the SUV to that of the car

$\tt \dfrac{mv}{1.5m\dfrac{2}{3}v }=1:1$

b. the ratio of the KE of the SUV to that of the car

$\tt \dfrac{1/2\times mv^2}{1/2\times 1.5m\times (\dfrac{2}{3}v )^2}=\dfrac{mv^2}{1.5m\dfrac{4}{6}v^2 }=1:1$

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A wire 95.0 mm long is bent in a right angle such that the wire starts at the origin and goes in a straight line to x = 40.0 mm

qaws [65]

Answer:

0.1040512455 N

$36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

0.05925 N

$29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

Explanation:

I = Current

B = Magnetic field

Separation between end points is

$l=\sqrt{40^2+55^2}\\\Rightarrow l=68.00735\ mm$

Effective force is given by

$F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N$

The force is 0.1040512455 N

$tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}$

The angle the force makes is given by

$\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

The direction is $36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

$F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N$

The force is 0.05925 N

$tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}$

$\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

The direction is $29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

4 0

3 years ago

A small rock is thrown vertically upward with a speed of 27.0 m/s from the edge of the roof of a 21.0-m-tall building. The rock

lbvjy [14]

Answer:

A. 33.77 m/s

B. 6.20 s

Explanation:

Frame of reference:

Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m

Initial velocity upwards v= 27 m/s

Part A. Using kinematics expression for velocities and distance:

$V_{final}^{2}=V_{initial}^{2}+2g(y_{final}-y_{initial})\\V_{f}^{2}=27^{2}-2*9.8(-21-0)=33.77 m/s$

Part B. Using Kinematics expression for distance, time and initial velocity

$y_{final}=y_{initial}+V_{initial}t+\frac{1}{2} g*t^{2}\\==> -0.5*9.8t^{2}+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s$

Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.

8 0

3 years ago

4. Many of the teachers think that if a student sleeps more on the night before the test, then they

Effectus [21]

Answer:

a sleep doesnt get to depict whether you get a good grade or not its what you know.

8 0

1 year ago

5) Why does the first hill/drop on a roller coaster have to be the highest? (think about

prohojiy [21]

5) Due to the conservation of energy / due to the presence of friction

6) The work done is 4000 J

7) The distance travelled by the object is 4 m

8) The potential energy of the bird is 2450 J

Explanation:

For a roller coaster in motion along the track, in absence of friction, the mechanical energy remains constant:

$E=U+K = const.$

where

U is the potential energy

K is the kinetic energy

At the beginning, the roller coaster is at rest, therefore its kinetic energy is zero and its mechanical energy is just equal to the potential energy:

$E=U=mgh$

where m is the mass of the roller coaster, g the acceleration of gravity, and h the height of the roller coaster above the ground at the first hill.

Since this amount of energy remains constant, the roller coaster cannot go to a higher hill: in fact, in that case it would reach a height $h'>h$ , but this is not possible, because it would mean that its mechanical energy would be $mgh' > E$ , larger than its mechanical energy, and since energy cannot be created (law of conservation of energy), the height of the next hills cannot be higher than the first one. Moreover, if we consider friction, then friction does work on the roller coaster in motion, and as a result, part of its mechanical energy is wasted (converted into thermal energy): therefore, the height of the following hills must be lower than the height of the first hill.

The work done when lifting an object is equal to the gravitational potential energy gained by the object:

$W=Fh$

where

F is the weight of the object

h is the change in height of the object

For the object in this problem, we have

F = 200 N (weight)

h = 20 m (change in height)

Substituting, we find the work done:

$W=(200)(20)=4000 J$

The work done when pushing an object in a direction parallel to the motion of the object is given by:

$W=Fd$

where

F is the magnitude of the force

d is the distance through which the object has travelled through

In this problem, we have:

W = 300 J (work done)

F = 75 N (magnitude of the force)

Solving the equation for d, we find the distance travelled:

$d=\frac{W}{F}=\frac{300}{75}=4 m$

The potential energy of an object is the energy possessed by an object due to its position in a gravitational field. Near the Earth's surface, it is given by

$PE=mgh$