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zalisa [80]
2 years ago
13

A skateboarder drops in off the top of one side of the half pipe shown below. She does not push off and starts from rest. She st

ands straight as she skates down one side and up the other. She expected to get to the top of other side but didn’t make it. The skateboarder recalled the law of conservation of energy from science and didn’t understand why she didn’t make it to the top of the other side.
Which of the following actions would help the skater reach the top of the other side of the half pipe?


Lubricate the wheels in order to reduce frictional force that is causing the skateboard to lose velocity and bend down while skating off the side and bottom of the half pipe.


Hold a weight in order to increase her mass and cause the skateboard to move with a greater velocity down the half pipe.


Sit on the skateboard in order to lower her center of mass and increase the potential energy of the skateboard.


Push off at the bottom of the half pipe in order to replace lost energy and add kinetic energy to the skateboard.
Physics
1 answer:
solong [7]2 years ago
4 0

Answer:

v

Explanation:

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In a second experiment, you decide to connect a string which has length L from a pivot to the side of block A (which has width d
Salsk061 [2.6K]

Answer:

The answer is in the explanation

Explanation:

A)

i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.

ii) The force on friction acting on the blocks is proportional to their mass, since mass of block B is less than block A, the force of friction acting on block B is also less. Hence, one might argue that block B will go farther along the table before coming to rest.

B) The equation of motion for block A is

m_{A}\frac{\mathrm{d} v}{\mathrm{d} t} = -m_{A}g\nu_{s}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t} = -\nu_{s}g \quad (1)

Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get

v(t) = C-\nu_{s}gt \quad (2)

Here, C is the constant of integration, which can be determined by using the initial condition

v(t=0) = v_{0}\Rightarrow C = v_{0} \quad (3)

Hence

v(t) = v_{0} - \nu_{s}gt \quad (4)

Block A will stop when its velocity will become zero,i.e

0 = v_{0}-\nu_{s}gT\Rightarrow T = \frac{v_{0}}{\nu_{s}g} \quad (5)

Going back to equation (4), we can write it as

\frac{\mathrm{d} x}{\mathrm{d} t} = v_{0}-\nu_{s}gt\Rightarrow x(t) = v_{0}t-\nu_{s}g\frac{t^{2}}{2}+D \quad (6)

Here, x(t) is the distance travelled by the block and D is again a constant of integration which can be determined by imposing the initial condition

x(t=0) = 0\Rightarrow D = 0 \quad (7)

The distance travelled by block A before stopping is

x(t=T) = v_{0}T-\nu_{s}g\frac{T^{2}}{2} = v_{0}\frac{v_{0}}{\nu_{s}g}-\nu_{s}g\frac{v_{0}^{2}}{2\nu_{s}^{2}g^{2}} = \frac{v_{0}^{2}}{2\nu_{s}g} \quad (8)

C) We can see that the expression for the distance travelled for block A is independent of its mass, therefore if we do the calculation for block B we will get the same result. Hence the reasoning for Student A and Student B are both correct, the effect of having larger initial energy due to larger mass is cancelled out by the effect of larger frictional force due to larger mass.

D)

i) The block A is moving in a circle of radius L+\frac{d}{2} , centered at the pivot, this is the distance of pivot from the center of mass of the block (assuming the block has uniform mass density). Because of circular motion there must be a centripetal force acting on the block in the radial direction, that must be provided by the tension in the string. Hence

T = \frac{m_{A}v^{2}}{L+\frac{d}{2}} \quad (9)

The speed of the block decreases with time due to friction, hence the speed of the block is maximum at the beginning of the motion, therfore the maximum tension is

T_{max} = \frac{m_{A}v_{0}^{2}}{L+\frac{d}{2}} \quad (10)

ii) The forces acting on the block are

a) Tension: Acting in the radially inwards direction, hence it is always perpendicular to the velocity of the block, therefore it does not change the speed of the block.

b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.

The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).

8 0
3 years ago
ou have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,
Dimas [21]

So far, since you moved into the apartment until the end of this much of the story, you haven't done ANY work on the dresser yet.

I'll admit that you pushed, groaned and grunted, sweated and strained plenty.  You're physically and mentally exhausted, you're not interested in the dresser at the moment, and right now you just want to snappa cappa brew, crash on the couch, and watch cartoons on TV.  But if you've done your Physics homework, you know you haven't technically done any <u><em>work</em></u> yet.

In Physics, "Work" is the product of Force times Distance.

Since the dresser hasn't budged yet, the Distahce is zero.  So no matter how great the Force may be, it's multiplied by zero, so the <em>Work is zero</em>.

5 0
3 years ago
The gravitational force, F, between an object and the Earth is inversely proportional to the square of the distance from the obj
andreev551 [17]
<h2>Weight of astronaut 2450 miles above the Earth is 80.38 pounds</h2>

Explanation:

Given that gravitational force, F, between an object and the Earth is inversely proportional to the square of the distance from the object and the center of the Earth.

             F=\frac{k}{r^2}

Where F is gravitational force  between an object and the Earth, r is the distance from the object and the center of the Earth and k is a constant.

Radius of Earth = 4000 miles

In case 1 an astronaut weighs 209 pounds on the surface of the Earth,

              209=\frac{k}{4000^2}\\\\k=3.344\times 10^9

Now we need to find weight of astronaut 2450 miles above the Earth    

              r = 4000 + 2450 = 6450 miles

               F=\frac{k}{6450^2}\\\\F=\frac{3.344\times 10^9}{6450^2}=80.38pounds

Weight of astronaut 2450 miles above the Earth is 80.38 pounds  

3 0
3 years ago
A train travels 120 km in 2 hours and 30 minutes. What is its average speed
marissa [1.9K]

Answer:

48 kmph

Explanation:

The question is asking you the kmph and that means the amount of kilometers travelled divided by the amount of hours. And if you want to think of it in a formula, SPEED = DISTANCE ÷ TIME

In this case, the amount of hours is 2.5 and the speed is 120km so 120/2.5 is 48

so the answer is 48 kmph

5 0
2 years ago
Read 2 more answers
HELP PLS WHAT IS Y COMPONENT
jeka94

Answer:

11.5

Explanation:

you do 12(sin 73.3) because its on the y you do sin

6 0
3 years ago
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