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3 years ago

A stone is dropped from the edge of the roof. (A) how long does it take to fall 4.9m

1 answer:

5 0

Since the stone is being dropped, you know that it is in free fall. That means you can use your kinematics equations, since acceleration is free fall is a constant 9.8 m/s^2 down.

Looking through the kinematics equations, you want to use one that lets you find t, time, while using variables that we already know the values of. Notice that in the equation:
$x_f = x_{0} + v_0 t + \frac{1}{2} a t^2$

Say that the rock starts at point x=0. That means the initial value of x, the position, is $x_0$ = 0m, and the final position is $x_f$ = -4.9m. We are also told that the initial velocity, $v_0$ = 0, assuming it's being dropped from rest, and a = -9.8 m/s^2. Plug these numbers in and solve for t:
$x_f = x_0 + v_0 t + \frac{1}{2} a t^2\\ x_f - x_0 = \frac{1}{2} a t^2\\ -4.9 - 0 = \frac{1}{2} a t^2\\ -9.8 = a t^2\\ -9.8 = -9.8 t^2\\ 1 = t^2\\ t = 1 \: second$

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Answer: t = 1 s

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Special relativity can be used to study an object in which frame of reference?

Tresset [83]

Special relativity led the path for general relativity; special relativity is in a sense a special application of the rules of general relativity. While general relativity is in position to tackle all of these problems, special relativity can tackle only problems in inertial frames. Inertial frame means that the frame of reference is inot accelerating. So, we disqualify answers A and D. However, remember that moving in a circle means that there is an acceleration, the centrifugal one, even if the speed does not change. Hence C is also incorrect.
The correct answer is B, since if there is no change in velocity, the frame does not accelerate and it is inertial.

3 0

3 years ago

Read 2 more answers

6. A 50 N block is raised 2 m. If the net work done on the block is 50 J, what is the applied force on the block?

geniusboy [140]

Answer:

F = 75[J]

Explanation:

We know that work is defined as the product of force by distance.

In this way we have two forces, the weight of the block down, and the force that bring about the block to rise.

$W = -(F_{weight*d})+(F_{upward}*d)$

where:

W = work = 50 [J]

d = distance = 2 [m]

Fweight = 50 [N]

Fupward [N]

Now replacing:

$50=-(50*2)+(F_{upward}*2)\\50+100=F_{upward}*2\\F_{upward}=150/2\\F_{upward}=75[J]$

5 0

3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0

3 years ago

A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subj

KATRIN_1 [288]

Answer:

$\Delta t = 8 s$

Explanation:

As we know that the angular acceleration of the wheel due to friction is constant

so we can use kinematics

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

so we have

$(65 \times 2\pi) = (2\pi \times 9)(10) + \frac{1}{2}(\alpha)(10^2)$

$130\pi = 180\pi + 50 \alpha$

$\alpha = -\pi rad/s^2$

now time required to completely stop the wheel is given as

$\omega_f = \omega_i + \alpha t$

$0 = (2\pi \times 9) + (-\pi) t$

$t = 18 s$

now time required to stop the wheel is given as

$\Delta t = 18 - 10$

$\Delta t = 8 s$

6 0

3 years ago

A Porsche sports car can accelerate at 8.8 m/s^2. Determine its acceleration in km/h^2.

finlep [7]

Answer:

The acceleration expressed in the new units is $114.048 Km/h^{2}$

Explanation:

To convert from $m/s^{2}$ to $Km/h^{2}$ it is necessary to remember that there are 1000 meters in 1 kilometer and 3600 seconds in 1 hour:

Then by means of a rule of three it is get:

$8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{3600s}{1h})^{2}$

$8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{12960000s^{2}}{1h^{2}})$

Hence, the units of meters and seconds will cancel. Notice the importance of square the ratio 3600s/1h, so that way they can match with the other units:

$114.048 Km/h^2$

So the acceleration expressed in the new units is $114.048 Km/h^2$ .

6 0

3 years ago