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Pavlova-9 [17]
3 years ago
5

An electromagnetic wave traveling through a vacuum has a wavelength of 1.5 × 10-1 meter. What is the period of this wave?

Physics
1 answer:
LiRa [457]3 years ago
8 0

Answer: (1) 5.0 x 10^ -10 s

Explanation:

1.5 x 10^-1 m / 3 x 10^8 = .5 x 10^-9

= 5 x 10^-10

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When it's at its highest temperature
4 0
3 years ago
Copper wire is 1. 9 mm in diameter and carries a current of 20 a. What is the electric field stregnth inside this wire?
Tju [1.3M]

The electric field strength will be 0.6252 V/m. It is the strength at which the field is created by charges.

<h3>What is electric file strength?</h3>

The electric field strength is defined as the ratio of electric force and charge.

The electric field strength is found as;

\rm E = \frac{I \rho }{A} \\\\ \rm E = \frac{20  \times 1.68 \times 10^{-8} }{ (0.6385 \times 10^{-6}} \\\\ E= 0.5262 \  V/m

Hence, the electric field strength will be 0.6252 V/m.

To learn more about the electric field strength, refer to the link;

brainly.com/question/4264413

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3 0
2 years ago
Calculate the upthrust aciting on a body if its
Zepler [3.9K]

Answer:

As a body moving upward

T=real weight + apparent weight

T=550+490

T=1040

hope u will get the answer:)

Explanation:

7 0
3 years ago
What is the force required to accelerate a 4 kg rock from 3 m/s to 15 m/s in seconds?
Lera25 [3.4K]

Answer:

16 Newtons

Explanation:

We will use the equation F=ma

First, find the acceleration using the givens.

a=∆v/∆t

a=15-3/3

a=12/3

a=4 m/s^2

Next, plug the acceleration and mass into the equation.

F=ma

F=4kg(4m/s^2)

F=16 Newtons

4 0
4 years ago
A car that weighs 1.0 x 10^4 N is initially moving at a speed of 38 km/h when the brakes are applied and the car is brought to a
hram777 [196]

Answer:

Part a) Force on car = 2833.84 Newtons

Part b) Time to stop the car = 3.8 seconds

Part c) Factor for stopping distance is 4.

Part d) Factor for stopping time is 1.

Explanation:

The deceleration produced when the car is brought to rest in 20 meters can be found by third equation of kinematics as

v^2=u^2+2as

where

v = final speed of the car ( = 0 in our case since the car stops)

u = initial speed of the car = 38 km/hr =\frac{38\times 1000}{3600}=10.56m/s

a = deceleration produces

s = distance in which the car stops

Applying the given values we get

0^2=10.56^2+2\times a\times 20\\\\a=\frac{0-10.56^2}{2\times 20}\\\\\therefore a=-2.78m/s^2

Now the force can be obtained using newton's second law as

Force=\frac{Weight}{g}\times a

Applying values we get

Force=\frac{1.0\times 10^4}{9.81}\times -2.78\\\\\therefore F=-2833.84Newtons

The negative direction indicates that the force is opposite to the motion of the object.

Part b)

The time required to stop the car can be found using the first equation of kinematics as

v=u+at with symbols having the same meanings

Applying values we get

0=10.56-2.78\times t\\\\\therefore t=\frac{10.56}{2.78}=3.8seconds

Part c)

From the developed relation of stopping distance we can see that the for same force( Same acceleration) the stopping distance is proportional to the square of the initial speed thus doubling the initial speed increases the stopping distance 4 times.

Part d)

From the relation of stopping time and the initial speed we can see that the stopping distance is proportional initial speed thus if we double the initial speed the stopping time also doubles.

8 0
3 years ago
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