All categories

3 years ago

A body travels 30m in 5s, 45m in 7s and then 65m in the last 5s. Find the average speed of the body

1 answer:

8 0

We can find the average speed of the body by finding the total distance covered, and then dividing it by the total time of the motion.

The total distance covered is:
$S=30 m + 45m+65m=140 m$

while the total time of the motion is
$t=5 s+7s+5 s=17 s$

So, the average speed of the body is:
$v= \frac{S}{t}= \frac{140 m}{17 s}=8.24 m/s$

You might be interested in

List down different layers of the sun. Rank these layers based on their distance from the sun’s center

vfiekz [6]

Answer:

Core

Radiative zone

Convective zone

Photosphere

Chromosphere

Transient region

Corona

Ranks of layers based on their distance from the sun’s center

1st-corona

2nd-Transient region

3rd-chromosphere

4th-Photosphere

5th-convective zone

6th-radiative zone

7th-core

6 0

3 years ago

Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of t

Lelu [443]

Answer:

c) At a distance greater than r

Explanation:

If G= Gravitational constant

M= Mass of earth

r= distance from earth center

then orbital speed is ;

v = $\sqrt{\frac{GM}{r} }$

==> v²=GM/r

If speed of first satellite = V₁

==> V₁² = GM/r

==> r = GM/V₁²

If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.

So our answer is c

5 0

3 years ago

A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced whe

gladu [14]

Answer:

x = 0.9 m

Explanation:

For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive

∑ τ = 0

60 1.5 - 78 1.5 + 30 x = 0

where x is measured from the left side of the fulcrum

90 - 117 + 30 x = 0

x = 27/30

x = 0.9 m

In summary the center of mass is on the side of the lightest weight x = 0.9 m

4 0

3 years ago

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

Irina-Kira [14]

Answer:

part (a) $\alpha\ =\ -2.5\ rad/s^2$

part (b) N = 79.61 rev

part (c) $\tau\ =\ 23.54\ Nm$

Explanation:

Given,

Initial speed of the wheel = $w_o\ =\ 50.0\ rad/s$
total time taken = t = 20.0 sec

part (a)

Let $\alpha$ be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

$\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2$

part (b)

Let $\theta$ be the total angular displacement of the wheel from initial position till the rest.

$\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad$

We know, 1 revolution = $2\pi$ rad

Let N be the number of revolution covered by the wheel.

$\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev$

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

Mass of the pole = m = 4 kg
Length of the pole = L = 2.5 m
Angle of the pole with the horizontal axis = $\theta\ =\ 60^o$

Now the center of mass of the pole = $d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m$

Weight component of the pole perpendicular to the center of mass = $F\ =\ mgcos\theta$

$\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm$

3 0

3 years ago

How will you describe the magnetic field around a current-carrying coil?

Gnesinka [82]

Well, in the center of the loop the field is more concentrated, and as it moves out it gets weaker. And when you stack more loops, it gets more concentrated, making a field called a solenoid.

8 0

3 years ago