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Sergeeva-Olga [200]

3 years ago

13

During the last shot of the game, the basketball goes from rest to 15 m/s and reaches the backboard in 0.41 s. What was the acce

leration of the ball?

2 answers:

Yuki888 [10]3 years ago

4 0

Answer:

The acceleration is 36.5853659

Explanation:

Acceleration is the change in velocity over the amount of time so for this it would be 15ms/.41s so the acceleration would be 36.5853659 $m/s^{2}$

Mrrafil [7]3 years ago

3 0

Answer:

When you climb, earth exerts gravitational force on pack in downward direction(pointing towards the center of earth).

In order to climb, you need to work against work done by gravity on the pack.

Hence work done by you = work done by gravity on pack

= Force x displacement = 70 x 30 = 2100 J.

So you need to do 2100 joules of work to lift your pack.

Power is the rate of work done.

Therefore power = work done by you/time(in seconds)

= 2100/600 =3.5 watts

Explanation:

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A child goes down the slide,starting from rest. If the length of the slide is 2m and it takes the child 3 seconds to go down the

lapo4ka [179]

Answer:

$0.44 m/s^2$

Explanation:

We have the following data:

- distance covered by the child: d = 2 m (length of the slide)

- time taken to cover this distance: t = 3 s

- initial velocity of the child: 0 m/s (he starts from rest)

So we can find the acceleration by using the equation:

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

Where a is the acceleration.

Substituting the values and solving for a,

$a=\frac{2d}{t^2}=\frac{2(2)}{3^2}=0.44 m/s^2$

3 0

3 years ago

You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon

Aleks [24]

Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

$u_{1}$ = 0m/s

$t_{1}$ = t

$u_{2}$ = 43.12m/s

$t_{2}$ = (t-2.2)s

The distance by the first balloon is

$D = u_{1} t_{1} + \frac{1}{2} at_{1}^2$

where

a = 9.8m/s2

Inputting the values

$D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2$

The distance traveled by the second balloon

$D = u_{2} t_{2} + \frac{1}{2} at_{2}^2$

Inputting the values

$D = (43.12)(t-2.2) + \frac{1}{2} (9.8)(t-2.2)^2$

simplifying

$D = 4.9t^2 + 21.56t -71.148$

Substituting D of the first balloon into the D of the second balloon and solving

$4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s$

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

$D = 4.9(3.3)^2\\ D = 53.361 m$

7 0

3 years ago

Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to

jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

brainly.com/question/12337396

#SPJ4

7 0

2 years ago

A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b

nekit [7.7K]

<h2>Answer:</h2>

In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:

$p(t)=v(t)i(t)$

So the average power is:

$P=\frac{1}{T}\intop_{0}^{T}p(t)dt$

But:

$v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)$

So:

$P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}$

$P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}$

In terms of RMS values:

$V_{RMS}=V=\frac{v_{m}}{\sqrt{2}} \\ \\ I_{RMS}=I=\frac{i_{m}}{\sqrt{2}} \\ \\ Then: \\ \\ P=VI$

7 0

3 years ago

Why are the coldest places on earth found near the poles

Nata [24]

Polar regions do not receive direct sunlight during the winter months due to the tilt in the Earth's<span> axis. Hence, polar regions can get very cold. Antarctica is the </span>coldest place on Earth. <span>The </span>coldest places on Earth<span> tend to be located </span>near the poles<span>. Hope this answers the question.</span>

6 0

3 years ago