All categories

3 years ago

What type of em wave travels the fastest?

Physics

1 answer:

nignag [31]3 years ago

5 0

I think Gamma Rays

Formula=3 ×× 10^8 m/s

You might be interested in

A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

Harman [31]

Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

coefficient of kinetic friction, $\mu_k=0.4$

So, the energy dissipated during the skidding of car:

Frictional force:

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

$f=5880\ N$

Now from the work-energy equivalence:

$E=f.d$

$E=5880\times 15$

$E=88200\ J$ is the dissipated energy.

3 0

3 years ago

A speeding motorist traveling with velocity Vm is spotted by a police car. The police car is initially at rest, but the instant

Svetlanka [38]

Answer:

Explanation:

Given

Let us suppose police car and motorist travel in straight line and police car catches motorist after s distance

Distance travel by motorist

$s=v_mt$ ----1

Distance traveled by Police car

$s=ut+\frac{at^2}{2}$

$s=0+\frac{a_pt^2}{2}$

$s=\frac{a_pt^2}{2}$ ----2

from 1 & 2 we get

$t=\frac{2v_m}{a_p}$

(a)Velocity of Police car after t sec

$v=u+a_pt$

$v=0+a_p\times \frac{2v_m}{a_p}$

$v=2v_m$

(b)time taken by police car is

$t=\frac{2v_m}{a_p}$

(c)Distance travel by police car $=\frac{2v_m^2}{a_p}$

7 0

3 years ago

A 50 kg student climbs 3m to the top of a set of stairs. Calculate the change in the student’s gravitational potential energy fr

erastova [34]

Gpe = mgh
gpe = 50*10*3
gpe = 1500 J

7 0

3 years ago

Convert 3.45inches into km

Katyanochek1 [597]

Answer:

can someone please answer this i need this for a mastery test aswell

Explanation:

it would be very appreciated

8 0

3 years ago

Read 2 more answers

A football punter accelerates a .55 kg football

Ronch [10]

Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the change in momentum of the football

$\Delta t=0.25 s$ is the time elapsed

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.55 kg is the mass of the football

u = 0 is the initial velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we find the force exerted on the ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N$

5 0

4 years ago