Which type of electromagnetic radiation is useful in communications technology

When an object accelerates, what about its motion changes? Question 1 options: Speed must change, but not velocity. Both speed a

QveST [7]

Answer:

The velocity must change but not speed.

Explanation:

Velocity is defined as the displacement by time. Whereas speed is expressed as the distance between two successive positions of the body to the time interval it took to travel.

Velocity, V = D / t m/s

Speed, s = d /t m/s

Velocity is a vector quantity that has a magnitude and direction.
The speed is a scalar quantity having only the magnitude.
At any instant of time, the magnitude of the velocity is always equal to the magnitude of the speed. The magnitude of velocity, |v | = magnitude of speed, |v |. The magnitude is always positive
The acceleration of a body is defined as the rate of change of velocity to time.

a = (v - u) / t m/s²

If a body is accelerating, It varies its velocity with respect to time.
In case of uniform circular motion, the speed remains constant, but the velocity changes continuously.

So, in the case of circular motion if an object accelerates, velocity must change but speed remains constant.

5 0

4 years ago

A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a.

Anna007 [38]

Answer:

$\mu_k = \frac{2(vt - L)}{(g + a) t^2}$

Explanation:

As we know that backpack is kicked on the rough floor with speed "v"

So here as per force equation in vertical direction we know that

$N - mg = ma$

so normal force on the block is given as

$N = mg + ma$

now the magnitude of kinetic friction on the block is given as

$F_f = \mu N$

$F_f = \mu (mg + ma)$

now when bag is sliding on the floor then net deceleration of the block due to friction is given as

$a = - \frac{F_f}{m}$

$a = -\mu_k(g + a)$

now we know that bag hits the opposite wall at L distance away in time t

so we have

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

$L = vt - \frac{1}{2}(\mu_k)(g + a) t^2$

$\mu_k = \frac{2(vt - L)}{(g + a) t^2}$

8 0

3 years ago

Tìm chỉ số của A1 và A2

ziro4ka [17]

Answer:

A1

Explanation:

6 0

3 years ago

An elevator filled with passengers has a mass of 1700 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s^{2

erica [24]

(a) 18717 N

Newton's second law in this situation can be written as:

$\sum F = T-W = ma$ (1)

where

T is the tension in the cable, pointing upward

W is the weight of the elevator+passengers, pointing downward

m is the mass of the elevator+passengers (1700 kg)

a is the acceleration of the system (1.20 m/s^2, upward)

The weight is equal to the product between the mass, m, and the gravitational acceleration, g:

$W=mg=(1700 kg)(9.81 m/s^2)=16,677 N$

So now we can solve eq.(1) to find T, the tension in the cable:

$T=W+ma=16,677 N +(1700 kg)(1.20 m/s^2)=18,717 N$

(b) 16677 N

In this situation, the elevator is moving with constant velocity: this means that its acceleration is zero,

a = 0

So Newton's second law becomes

$\sum F = T-W = 0$

and so we find

$T=W=16,677 N$

(c) 15657 N

During the deceleration phase, Newton's second law can be written as:

$\sum F = T-W = ma$ (1)

Where the acceleration here points downward (because the elevator is decelerating), as the weight W, so we can write it as a negative number:

a = -0.600 m/s^2

we can solve the equation to find T, the tension in the cable:

$T=W+ma=16,677 N +(1700 kg)(-0.600 m/s^2)=15,657 N$

(d) 19.35 m, 0 m/s

Distance covered during the first part of the motion; we know that

u = 0 is the initial velocity

a = 1.20 m/s^2 is the acceleration

t = 1.50 s is the time

So the distance covered is given by

$d_1=ut + \frac{1}{2}at^2 = (0)(1.50 s)+\frac{1}{2}(1.20 m/s^2)(1.50 s)^2=1.35 m$

and the final velocity after this phase is

$v_1=u+at=0+(1.20 m/s^2)(1.50 s)=1.8 m/s$

During the 2nd part of the motion, the elevator moves at constant speed of 1.8 m/s for t=8.50 s, so the distance covered here is

$d_2 = v_1 t =(1.8 m/s)(8.50 s)=15.3 m$

Finally, in the third part the elevator decelerates at a = -0.600 m/s^2 for t = 3.00 s. So, the distance covered here is

$d_3 = v_1 t + \frac{1}{2}at^2=(1.8 m/s)(3.00 s) + \frac{1}{2}(-0.600 m/s^2)(3.00 s)^2=2.7 m$

and the final velocity is

$v_3 = v_1 +at = 1.8 m/s +(-0.600 m/s^2)(3.00 s)=0$

and the total distance covered is

$d=d_1 +d_2+d_3=1.35 m+15.30 m+2.70 m=19.35 m$

3 0

3 years ago

A bicyclist rides 1.59 km due east, while the resistive force from the air has a magnitude of 7.00 N and points due west. The ri

kramer

Answer:

The required work done by the resistive force = -22.3 kJ

Explanation:

We understand that the amount of work done for a force F and displacements is provided as the product of force multiplied by displacement, which can be represented as W = F.S

Now, when the cyclist travels east and returns west, the force and displacement directions will be opposite and they make an angle 180 degrees apart.

∴

W = -2 × 7.00 × 1590

W = -22260 J

W = -22.3 kJ

8 0

3 years ago