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ch4aika [34]
3 years ago
9

What is the shape of the trajectory that a charged particle follows in a uniform magnetic field?

Physics
1 answer:
Darya [45]3 years ago
5 0

Answer:

circle

Explanation:

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A 920-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,
GrogVix [38]

Answer:21.45 m/s

Explanation:

Given

Mass of sport car=920 kg

Mass of SUV=2300 kg

distance to which both car skid is 2.4 m

coefficient of friction (\mu)=0.8

Let u be the initial velocity of both car at the starting of skidding

and they finally come to zero velocity

v^2-u^2=2as

acceleration=\mu g=0.8\times 9.8=7.84 m/s^2

s=2.4 m

0-(u)^2=2\times (-7.84)\times 2.4

u=6.13 m/s

so before colliding sport car must be travelling at a speed of

920\times v=(920+2300)\times 6.13  (conserving momentum)

v=21.45 m/s

7 0
3 years ago
At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m2. A fan is 30 m from the loudspeak
aliina [53]

Explanation:

Below is an attachment containing the solution.

5 0
3 years ago
SP: Calculate the moment
ipn [44]

Answer:

Moment of the force is 20 N-m.

Explanation:

Given:

Force exerted by the person is, F=80\ N

Distance of application of force from the point about which moment is needed is, d=25\ cm=\frac{25}{100}\ m=0.25\ m

Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.

Therefore, the moment of the force about the end of the claw hammer is given as:

M=F\times d\\\\M=(80\ N)(0.25\ m)\\\\M=20\textrm{ N-m}

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.

6 0
3 years ago
After a massive-star supernova, what is left behind?.
scoray [572]

Answer: A <u>Nebula </u>is left behind. A spectacular explosion in which a star ejects most of its mass in a violently expanding cloud of debris.

Hope this helps!

8 0
2 years ago
in the derivation of the time period of a pendulum in electric field when considering the fbd of bob to find the g effective why
Neko [114]

Answer:

we learned that an object that is vibrating is acted upon by a restoring force. The restoring force causes the vibrating object to slow down as it moves away from the equilibrium position and to speed up as it approaches the equilibrium position. It is this restoring force that is responsible for the vibration. So what forces act upon a pendulum bob? And what is the restoring force for a pendulum? There are two dominant forces acting upon a pendulum bob at all times during the course of its motion. There is the force of gravity that acts downward upon the bob. It results from the Earth's mass attracting the mass of the bob. And there is a tension force acting upward and towards the pivot point of the pendulum. The tension force results from the string pulling upon the bob of the pendulum. In our discussion, we will ignore the influence of air resistance - a third force that always opposes the motion of the bob as it swings to and fro. The air resistance force is relatively weak compared to the two dominant forces.

The gravity force is highly predictable; it is always in the same direction (down) and always of the same magnitude - mass*9.8 N/kg. The tension force is considerably less predictable. Both its direction and its magnitude change as the bob swings to and fro. The direction of the tension force is always towards the pivot point. So as the bob swings to the left of its equilibrium position, the tension force is at an angle - directed upwards and to the right. And as the bob swings to the right of its equilibrium position, the tension is directed upwards and to the left. The diagram below depicts the direction of these two forces at five different positions over the course of the pendulum's path.

that's what I know so far

8 0
3 years ago
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