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ANEK [815]
3 years ago
13

Magnetism is a type ofA. forceB. energyC. gravityD. electricity​

Physics
2 answers:
erastovalidia [21]3 years ago
5 0

Answer:

A. force

Explanation:

Magnetism is a type of force.

Vlad1618 [11]3 years ago
4 0

Answer:

A. force

Hope this helps

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The nucleus of most atoms is composed of which of the following sub-atomic particles?
Blababa [14]

Answer:

protons (+ charge) & neutrons (neutral charge)

however protons has a positive charge so it determined what atom it is.

7 0
4 years ago
Suppose a 2.0kg bird is flying at a speed of of 1m/s. It’s kinetic energy would be what?
Ganezh [65]

As we know that the formula of kinetic energy will be

KE = \frac{1}{2} mv^2

now here we know that

m = 2 kg

v = 1 m/s

so from the above equation we have

KE = \frac{1}{2}(2)(1^2)

KE = 1 J

7 0
3 years ago
4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest
yarga [219]

All the answers are:

a) The time that will it take to rise to the highest point is 3.06 seconds.

b) The ball will rise to a height of 45.87 meters.

c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.

The time when the ball has 10 m/s downward is 1.02 seconds.

d) The displacement of the ball will be zero at 6.12 seconds.

e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.

f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.

g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Let's calculate the values for each case.

a) At the highest point, the final velocity is 0, so we can use the following equation.  

v_{f}=v_{i}-gt (1)

Where:

  • v(i) is the initial velocity
  • v(f) is the final velocity
  • g is the acceleration due to gravity (9.81 m/s²)

We know that v(i) = 30 m/s.

0=30-9.81t

Solve it for t:

t=3.06\: s

Hence, the time is 3.06 s.

b) At the highest point, the final velocity is 0, so we can use the following equation.  

v_{f}^{2}=v_{i}^{2}-2gh (2)

0=v_{i}^{2}-2gh

We know that the initial velocity is 30 m/s.

0=30^{2}-2gh

Solving it for h we have:  

h=\frac{30^{2}}{2*9.81}

h=45.87 \: m

Then, the height is 45.87 m.

c) Using equation (1) we can find the time (t).

10=30-(9.81t)

So, the time elapsed to get 10 m/s is:

t_{upward}=2.04\: s

We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.

t_{upward}=2.04+t  

t=1.02\: s

This is the time when the ball has 10 m/s downward.          

Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.

d) It will be when the ball returns to the ground.

t=2t_{upward}

t=2*3.06      

t=6.12\: s

The displacement will be zero after 6.12 s.  

e) Here we need to find the time when v(f) is 15 m/s

15=30-gt

t=\frac{15}{9.81}  

t=1.53\: s

The time when the v(f) is 15 m/s is 1.53 s.

f) Here, we need to find t when h = 45.87/2 m = 22.94 m

We can use the next equation:

[tex]h=v_{i}t-0.5gt^{2}/tex]

[tex]22.94=30t-0.5*9.81*t^{2}/tex]

Solving this quadratic equation, t will be:

[tex]t=0.90\: s/tex]

Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.

g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Learn more about vertical motion here:

brainly.com/question/13966860

I hope it helps you!

3 0
3 years ago
A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d
Aleks04 [339]
I’m pretty sure it’s b
8 0
3 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
4 years ago
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