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Morgarella [4.7K]

3 years ago

8

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1 answer:

slamgirl [31]3 years ago

8 0

Answer:

Speed at which it will reach the ground is given as

$v_f = 46.8 m/s$

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

$v_i = 15 m/s$

$g = - 9.81 m/s^2$

now when it will reach the ground then we have

$y = v_y t + \frac{1}{2} at^2$

so we have

$-100 = 15 t - \frac{1}{2}(-9.81) t^2$

$4.905 t^2 - 15 t - 100 = 0$

so we have

$t = 6.3 s$

Now speed of the object when it reaches the ground is given as

$v_f = v_i + at$

$v_f = -15 + (9.81)(6.3)$

$v_f = 46.8 m/s$

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Lisa [10]

Since the basketball and the tennis ball both travel to the same direction relative to the ground, the velocity of the basketball relative to the tennis ball is therefore the difference of their velocities.
0.5 m/s - 0.25 m/s = 0.25 m/s
Thus, the basketball travel for 0.25 m/s relative to the tennis ball.

8 0

3 years ago

Find the intensity of electromagnetic radiation at the surface of the sun (radius r=R=6.96×105kmr=R=6.96×105km). Ignore any scat

alisha [4.7K]

Answer:

$I = 4.46*10^{16}W/m^2$ .

Explanation:

Intensity $I$ of the electromagnetic radiation is given by

$I = \dfrac{P}{4\pi r^2},$

where $r$ is the distance from the EM source (the center of the sun, in our case), and $P$ is the power output of the sun and it has the value

$P = 3.9 *10^{26}W.$

Since the radius of the sun in meters is $r = 6.96*10^8km$ , the intensity $I$ of the electromagnetic radiation at the surface of the sun is

$I = \dfrac{3.9*10^{26}W}{4\pi (6.96*10^8m)^2}\\\\\boxed{ I = 4.46*10^{16}W/m^2}$

The intensity of the electromagnetic radiation at the surface of the sun is $I = 4.46*10^{16}W/m^2$ .

7 0

3 years ago

(ii) Describe how the acceleration of the train at time t = 100 s differs from the acceleration

quester [9]

Explanation:

Acceleration is the rate of change of velocity with time. When acceleration increases a body moves a faster velocity.

In the graph acceleration at time t= 100s is rapidly increasing.
At t = 20s, the acceleration of the body is getting started up.

A vehicle at time 100s will have a faster velocity compared to one at t = 20s

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3 years ago

Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered

Soloha48 [4]

Answer:

$\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}$

$\boxed{\sf Distance \ covered \ (s) = 100 \ m}$

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

$\sf (i) \ From \ 1^{st} \ equation \ of \ motion:$

$\sf \implies v = u + at$

$\sf \implies 20 = 0 + a(10)$

$\sf \implies 10a = 20$

$\sf \implies \frac{10a}{10} = \frac{20}{10}$

$\sf \implies a = 2 \: m/ {s}^{2}$

$\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:$

$\sf \implies s = ut + \frac{1}{2} a {t}^{2}$

$\sf \implies s = (0)(10) + \frac{1}{2} \times 2 \times {(10)}^{2}$

$\sf \implies s = \frac{1}{ \cancel{2}} \times \cancel{2} \times {(10)}^{2}$

$\sf \implies s = {10}^{2}$

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6 0

3 years ago

A truck initially traveling at a speed of 22 m/s increases at a constant rate of 2.4 m/s^2 for 3.2s. What is the total distance

FinnZ [79.3K]

Answer:

82.7 m

Explanation:

u= 22m/s

a= 2.4 m/s^2.

t= 3.2 secs

Therefore the distance travelled can be calculated as follows

S= ut + 1/2at^2

= 22 × 3.2 + 1/2 × 2.4 × 3.2^2

= 70.4 + 1/2×24.58

= 70.4 + 12.29

= 82.7 m

Hence the distance travelled by the truck is 82.7 m

6 0

2 years ago