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Valentin [98]
4 years ago
13

What has alternating electric and magnetic fields that travel in the form of a wave

Physics
1 answer:
Margaret [11]4 years ago
8 0
Electromagnetic radiation
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Consider a situation where the acceleration of an object is always directed perpendicular to its velocity. This means that
Bond [772]

Answer:

this situation would not be physically possible

7 0
3 years ago
What is the average power consumption in watts of an appliance that uses 5.00 KW.h of energy per day?
abruzzese [7]
(A) power = 0.208 kW = 208 watts
(B) energy = 6.6 x 10^{9} joules

Explanation:
energy consumed per day = 5 kWh
(a) find the power consumed in a day
1 day = 24 hours
power = \frac{energy}{time}
power = \frac{5}{24}
power = 0.208 kW = 208 watts

(b) find the energy consumed in a year
assuming it is not a leap year and number of days = 365 days
1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds
energy = power x time
energy = 208 x 31,536,000
energy = 6.6 x 10^{9} joules
3 0
3 years ago
4. How much mass is required to exert a force of 25 Newtons, accelerating at 5 m/s?
lions [1.4K]
F = ma
F/a = m
(25 N) / (5 m/s2) = m
4 kg = m
8 0
3 years ago
Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each mo
postnew [5]

Answer:

ucosφ=-v2cosθ2\\\\φ=cos^{-1} (\frac{-v2cosθ2}{cosφ} )

Explanation:

Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each moving at the same speed v when they collide at the origin of the coordinates and stick together. After the collision, the masses move at an angle −θ2 with respect to the +x axis at speed v2 .1. What was the angle φ?

from the principle of momentum

In a system of colliding bodies,we know that the total momentum before collision will equal to the total momentum after collision.

Take note that momentum is the product of mass and velocity

momentum before collision=momentum after collision

mass, m

u=initial velocity of the identical masses

v2=the common velocity after the collision

Note that the collision is inelastic , since they both moved with the same velocity

umcosφ+umcosφ=(m+m)v2cos−θ2

2mucosφ=2mv2cos−θ2

ucosφ=-v2cosθ2\\\\φ=cos^{-1} (\frac{-v2cosθ2}{cosφ} )

8 0
3 years ago
How much work is done if you push a box 150 meters with a force of 1.4 N
ss7ja [257]
The work done to push the box is equal to the product between the force and the distance through which the force is applied:
W=Fd
In our problem, the force is F=1.4 N and the distance covered is d=150 m, so the work done by pushing the box is
W=Fd=(1.4 N)(150 m)=210 J
6 0
3 years ago
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