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Crank
4 years ago
15

I’m completely lost, help please.

Physics
1 answer:
Lana71 [14]4 years ago
8 0

Relax.  You're not completely lost.

The car has to cover 600 km.  That's 600,000 meters.

If it could only move at 1 meter per second, it would take the car 600,000 seconds to cover the distance.  (almost a week !)

But it can move faster than that.  In fact, it can cover 80 meters every second.

How many times does it need to do 80 meters, in order to cover the whole 600,000 ?  That's how many seconds it'll take.

That's division.

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When he sees teachers encouraging other children to wait in the cafeteria until the first bell rings, Ian follows them. What typ
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This is observational learning because Ian observed that his peers waited in the cafeteria until the first bell rings. He decided to imitate them.
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3 years ago
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What would happen to a 100N gravitational force if both masses were doubled and the radius were
grin007 [14]

Answer:

They are the only joints that can do 360 degrees and rotate with their own axis. But, because of its free-moving, it is prone to any dislocation compared to other movable joints.

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An object is dropped from a platform100 ft high. Ignoring wind resistance, what will its speed be when it reaches the ground?
sweet-ann [11.9K]

Answer:

80 ft/s

Explanation:

Given:

Δy = 100 ft

v₀ = 0 ft/s

a = 32.2 ft/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 ft/s)² + 2 (32.2 ft/s²) (100 ft)

v = 80.2 ft/s

Rounded, the speed when it reaches the ground is 80 ft/s.

4 0
3 years ago
A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T.
Gekata [30.6K]

As per the question, the point charge is given as [q] = 6.8 C

The velocity of the charged particle [v] = 6.5*10^{4}\ m/s

The magnetic field [B] = 1.4 T

The angle made between magnetic field and velocity [\theta] = 15 degree.

We are asked to calculate the magnetic force experienced by the charged  particle.

The magnetic force experienced by the charged particle is calculated as -

Magnetic force \vec F=q(\ \vec V \times \vec B)

                        i.e F = =qVBsin\theta

                                   =6.8*6.5*10^{4}*1.4*sin15

                                   =61.88*10^{4}sin15

                                   =61.88*10^4*0.2588\ N

                                   =16.016*10^4\ N

                                   =1.6*10^5\ N

Hence, the force experienced by the charged particle is C i.e 1.6*10^5 N

                                   

             

4 0
3 years ago
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Write down the conservation of momentum?​
otez555 [7]
Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies.
Let,
m
A
​
= Mass of ball A
m
B
​
= Mass of ball B
u
A
​
= initial velocity of ball A
u
B
​
= initial velocity of ball B
v
A
​
= Velocity after the collision of ball A
v
B
​
= Velocity after the collision of ball B
F
ab
​
= Force exerted by A on B
F
ba
​
= Force exerted by B on A
Now,
Change in the momentum of A= momentum of A after the collision - the momentum of A before the collision
= m
A
​
v
A
​
−m
A
​
u
A
​

Rate of change of momentum A= Change in momentum of A/ time taken
=
t
m
A
​
v
A
​
−m
A
​
u
A
​

​

Force exerted by B on A (F
ba
​
);
F
ba
​
=
t
m
A
​
v
A
​
−m
A
​
u
A
​

​
........ [i]
In the same way,
Rate of change of momentum of B=
t
m
b
​
v
B
​
−m
B
​
u
B
​

​

Force exerted by A on B (F
ab
​
)=
F
ab
​
=
t
m
B
​
v
B
​
−m
B
​
u
B
​

​
.......... [ii]
Newton's third law of motion states that every action has an equal and opposite reaction, then,
F
a
​
b=−F
b
​
a [ ' -- ' sign is used to indicate that 1 object is moving in opposite direction after collision]

Using [i] and [ii] , we have
t
m
B
​
v
B
​
−m
B
​
u
B
​

​
=−
t
m
A
​
v
A
​
−m
A
​
u
A
​

​

m
B
​
v
B
​
−m
B
​
u
B
​
=−m
A
​
v
A
​
+m
A
​
u
A
​

Finally we get,
m
B
​
v
B
​
+m
A
​
v
A
​
=m
B
​
u
B
​
+m
A
​
u
A
​

This is the derivation of conservation of linear momentum.
5 0
3 years ago
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