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2 years ago

I NEED HELP I AM SO CONFUSED, WILL GIVE BRAIN.

Physics

1 answer:

MakcuM [25]2 years ago

5 0

Answer:

210

Explanation:

A ball rolls horizontally off the cliff at a speed of 30 m/s. It takes 7 seconds for the ball to hit the ground. What is the height of the cliff and the horizontal distance traveled by the ball?

S = (1/2)*9.8 m/s^2 * 7^2 = 240.1 m if the ball is very dense so air resistance, and therefore terminal velocity, can be ignored.

S = v * t = 30 m/s * 7 s = 210 m for the horizontal distance, again assuming negligible air resistance.

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What Determines the amount of TE of an object

Luden [163]

Answer:

To find the volume of a rectangular object, measure the length, width and height. Multiply the length times the width and multiply the result by the height. The result is the volume. Give the result in cubic units, such as cubic centimeters.

Explanation:

7 0

2 years ago

Read 2 more answers

A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s

Ede4ka [16]

When acceleration is constant, the average velocity is given by

$\bar v=\dfrac{v+v_0}2$

where $v$ and $v_0$ are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

$\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}$

where $x,x_0$ are the final/initial displacements, and $t,t_0$ are the final/initial times, respectively.

Take the car's starting position to be at $t_0=0\,\mathrm s$ . Then

$\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t$

So we have

$x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m$

You also could have first found the acceleration using the equation

$v=v_0+at$

then solve for $x$ via

$x=x_0+v_0t+\dfrac12at^2$

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of $a$ anyway.

7 0

2 years ago

What is the average acceleration of the particle between 0 seconds and 4 seconds? A. 0 meters/second2 B. 0.04 meters/second2 C.

lisov135 [29]

Answer

D. 0.25 meters/second2

Explanation

The average acceleration is the ratio of change in velocity to the change in time of travel.Taking in this case that the change of velocity is a unit, then Average acceleration is given by;

Aacc=Vf-Vi/Tf-Ti

where Vf=final velocity,Vi=initial velocity' Tf=final time, Ti=initial time

Vf-Vi=1m/s

Tf-Ti=4-0=4seconds

Avacc=1/4=0.25m/s2

6 0

2 years ago

You normally drive a 12-h trip at an average speed of 100 km/h . Today you are in a hurry. During the first two-thirds of the di

kherson [118]

Answer:

78 km/h

Explanation:

If I normally drive a 12 hour trip at an average speed of 100 km/h, my destination has a total distance of:

100 km/h · 12 h = 1,200 km

Today, I drive the first 2/3 of the distance at 116 km/h. Let's first calculate what 2/3 of the normal distance is.

1,200 * 2/3 = 800 km

I've driven 800 km already. I need to drive 400 km more to reach my final destination. I need to figure out my average speed during this last 1/3 of the distance.

To do this, I first need to calculate how much time I spent driving 116 km/h for the past 800 km.

116 km/1 h = 800 km/? h
800 = 116 · ?
? = 800/116
? = 6.89655172

I spent 6.89655172 hours driving during the first 2/3 of the distance.

Now, I need to subtract this value from 12 hours to find the remaining time I have left.

12 h - 6.89655172 h = 5.10344828 h

Using this remaining time and my remaining distance, I can calculate my average speed.

? km/1 hr = 400 km/5.10344828 h
5.10344828 · ? = 400
? = 400/5.10344828
? = 78.3783783148

My average speed during the last third of the distance is around 78 km/h.

8 0

2 years ago

Spiders may "tune" strands of their webs to give an enhanced response at frequencies corresponding to the frequencies at which d

Bingel [31]

Answer:

0.0000167283525619 N

Explanation:

$\rho$ = Density of silk = 1300 kg/m³

d = Diameter = 0.002 mm

r = Radius = 0.001 mm

l = Length = 16 cm

f = Frequency = 200 Hz

Mass of the string is

$m=\rho V\\\Rightarrow m=\rho Al\\\Rightarrow m=1300\times \pi (0.001\times 10^{-3})^2\times 0.16\\\Rightarrow m=6.5345127195\times 10^{-10}\ kg$

Frequency is given by

$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{m/l}}\\\Rightarrow T=4lf^2m\\\Rightarrow T=4\times 0.16\times 200^2\times 6.5345127195\times 10^{-10}\\\Rightarrow T=0.0000167283525619\ N$

The tension on the string is 0.0000167283525619 N

3 0

3 years ago