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3 years ago

Is radiation the transfer of energy in the form of particles

1 answer:

7 0

Radiation is the emission of energy as electromagnetic waves or as moving subatomic particles, especially high-energy particles that cause ionization.
Or, Radiation is the energy transmitted by heat, light, electricity, etc. However, your question term is convection.

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A car accelerates at a constant rate from 12 m/s to 27 m/s while it travels 125 m. How long does it take to achieve this speed?

MArishka [77]

Answer:

6.41 s

Explanation:

Under constant acceleration we know that

average velocity × time taken = displacement

$s=\frac{u+v}{2}t$

$125=\frac{27+12}{2}t$

t = 6.41 s

The proof of used equation is given in the attachment.

3 0

3 years ago

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 30 m/s2 for 35 s , then ru

Gre4nikov [31]

Answer:

a) $h_m=74625\ m$

b) $t=265.55\ s$

Explanation:

Given:

mass of rocket, $m_r=200\ kg$

mass of fuel, $m_f=100\ kg$

acceleration of the rocket consuming fuel, $a=30\ m.s^{-2}$

time after which the fuel exhaust, $t_f=35\ s$

During the phase of fuel exhaustion:

velocity attained by the rocket just as the fuel ends:

$v_f=u+a.t_f$

where:

$u=$ initial velocity of the rocket = 0

$v_f=0+30\times 35$

$v_f=1050\ m.s^{-1}$ this will be the initial velocity for the phase of ascending of the rocket's height under the influence of gravity.

height at which the fuel finishes:

$v_f^2=u^2+2a.h_f$

$1050^2=0^2+2\times 30\times h_f$

$h_f=18375\ m$

During the phase of ascend in height of rocket after the fuel is over:

Time taken to reach the top height after the fuel is over:

$v=v_f+g.t'$

at top v = (final velocity during this course of motion )= 0 $m.s^{-1}$

$0=1050-9.8\times t'$

$t'=107.1429\ s$

Height ascended by the rocket after the fuel is over:

$v^2=v_f^2+2g.h'$

at the top height the velocity is zero

$0^2=1050^2-2\times 9.8\times h'$ (-ve sign denotes that the direction of motion is opposite to that of acceleration)

$h'=56250\ m$

Therefore the maximum altitude attained by the rocket:

$h_m=h_f+h'$

$h_m=18375+56250$

$h_m=74625\ m$

time taken by the rocket to fall back to the earth:

$h_m=v.t_m+\frac{1}{2} g.t_m^2$

where:

$v=$ initial velocity of the rocket during the course of free fall from the top height.

$74625=0+4.9\times t_m^2$

$t_m=123.41\ s$

Now the total time for which the rocket is in the air:

$t=t_f+t'+t_m$

$t=35+107.1429+123.41$

$t=265.55\ s$

4 0

3 years ago

Which two statements about an electric motor are true?

allochka39001 [22]

Options (b) and (d) are correct about an electric motor.

An electric motor is a device which coverts electrical energy into mechanical energy.It works on the principle that when a current carrying coil is placed in a magnetic field, it experiences torque. Because of that torque, the coil rotates and thus the electrical energy gets converted into mechanical (motion) energy.

7 0

3 years ago

Read 2 more answers

In Ancient Greece, athletes competing in the long jump used handheld weights called halteres to lengthen their jumps. You are a

katovenus [111]

The halter add the distance to the jump in meters is 0.55 m.

<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.

The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.

Components of velocity in x and y direction are

Vx = 10.3 cos 22.8 = 9.5 m/s

Vy = 10.3 sin 22.8 = 4 m/s

Maximum Range of athlete achieved using halter is given by

R = u²sin2θ /g

where, u = initial velocity, θ is the angle of projection and g is the gravitational acceleration.

Substituting the values, we get

R = (10.3)² sin(2 x 22.8 °) / 2 x 9.81

R = 7.75m

At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.

The momentum is conserved in this situation,

(M+2m)Vxo =MVx'

Vx' = (M+2m)/M x Vxo'

Change in x component of velocity ΔVx = Vx' -Vxo

Vxo = 2m/M x Vx

Vxo = 2 x 5.5 /78 x 9.5

Vxo = 1.34 s

Maximum height gained when final velocity is zero

Vy = 0 = Vyo -gt

time t = Vyo/g = 4/9.8 = 0.41s'

Increase in range by using of halters is

ΔR = ΔVx' x t

ΔR = 1.34 x 0.41

ΔR =0.55m

Thus, the halter add the distance to the jump in meters is 0.55 m.

Learn more about projectile.

brainly.com/question/11422992

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3 0

2 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: