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3 years ago

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Physics

1 answer:

omeli [17]3 years ago

7 0

Answer:

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When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it fro

alexandr402 [8]

Answer:

Explanation:

Given

When we drop an object from height , suppose h

it takes time T

using equation of motion

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

where

$h=displacement$

$u=initial\ velocity$

$a=acceleration$

$t=time$

here $u=0$ because it dropped from a certain height

$h=\frac{1}{2}gT^2$

$T=\sqrt{\frac{2h}{g}}$

When height is increases to three times of original height

i.e. $h'=3 h$

then time period becomes

$T'=\sqrt{\frac{2\times 3h}{g}}$

$T'=\sqrt{3}T$

5 0

2 years ago

What is the momentum of an 80 kg runner moving at the speed of 2.5 m/s? Use the

Vladimir79 [104]

Answer:

200 kgm/s

Explanation:

momentum = mass x velocity

8 0

2 years ago

Suppose a 1300 kg car is traveling around a circular curve in a road at a constant

Scrat [10]

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :

$F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N$

So, the force has a magnitude of 4212 N

4 0

2 years ago

Two identical point charges are 3.00 cm apart. find the charge on each of them if the force or repulsion is 4.00 x 10^-7. (Use C

DanielleElmas [232]

Answer:

Charge on each is 2 x 10⁻¹⁰.

Explanation:

We know that Force between two point charges is given b the Coulomb's law as:

F = kq₁q₂/r^2

k = 9 x 10^9

r = 3.00 cm

= 0.03 m

q₁ = q₂

F = 4.00 x 10^-7

Rearranging the formula, we get:

F = k q²/r²

q² = Fr²/k

q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)

q² = 4 x 10⁻²⁰

q = 2 x 10⁻¹⁰

As there is force of repulsion between the charges, the charges must be both positive or both negative.

3 0

3 years ago

Two boats - Boat A and Boat B - are anchored a distance of 24 meters apart. The incoming water waves force the boats to oscillat

ozzi

Answer:

wavelength = 24 m

Period = 10 s

f = 0.1 Hz

Amplitude = 4 m

Explanation:

Wavelength:

Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:

wavelength = 24 m

Period:

The period is given as:

$Period = \frac{time}{no.\ of\ cycles} \\\\Period = \frac{10\ s}{1}\\\\$

Period = 10 s

Frequency:

The frequency is given as:

$f = \frac{1}{time\ period}\\\\f = \frac{1}{10\ s}\\\\$

f = 0.1 Hz

Amplitude:

Amplitude will be half the distance between extreme points, that is, crest and trough:

Amplitude = 8 m/2

Amplitude = 4 m

5 0

2 years ago