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3 years ago

A grapefruit falls from a tree and hits the ground 0.76 s later.

Physics

1 answer:

Tomtit [17]3 years ago

7 0

Answer:

2.8 m 7.4 m/s

Explanation:

write all the values then use the equations of motion to find the distance and speed. please see attached photo

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A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

stealth61 [152]

Answer :

(a) The acceleration of the car is, $-2m/s^2$

(b) The distance covered by the car is, 150 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

The acceleration of the car is, $-2m/s^2$

By the 2nd equation of motion,

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

where,

s = distance covered by the car = ?

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

By solving the term, we get:

$s=150m$

The distance covered by the car is, 150 m

8 0

2 years ago

A lamp draws a current of 0.1 A when it is connected to a 122-V source. (

Setler79 [48]

The power of the lamp would be calculated with the equation of ohm laws. P = U x I = 122V x 0.1A = 12.2W

4 0

2 years ago

The odometer of a car changes from 1048 km to 1096 km in 40

Makovka662 [10]

Answer:

20m/s

Explanation:

it covers 20 metres in a second

3 0

1 year ago

A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st

Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

$\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r$

where,

$\tau$ = shear stress at a distance 'r' from the center

T = is the applied torque

$I_{p}$ = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

$\tau _{max}=\frac{4}{3}\times \frac{V}{A}$

Applying values we get

$\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa$

3 0

2 years ago

What is the motion of the particles in this kind of wave?

Musya8 [376]

The answer to this question is B I think

7 0

2 years ago

Read 2 more answers