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sweet-ann [11.9K]
3 years ago
7

A magnetic field would be produced by a beam of

Physics
1 answer:
kumpel [21]3 years ago
4 0
3 protons should be your answer
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Please answer the following question
aleksklad [387]

Answer:

1. Light Bulb

2. A switch

3. A battery

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3 years ago
A block of mass 2 kg slides down a frictionless ramp of length 1.3 m tilted at an angle 25o to the horizontal. At the bottom of
marin [14]

Answer:

Diagrams in pictures

Explanation:

Using energy I can get

m g h = 1/2 m v^2

So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.

(H= 1,3m sin25)

V=2,32m/s

V= a t

And

X= v t +1/2 a t^2

Knowing v=2,32 m/s and x= 1,3 m

I can get

a= 6,21m/s2

F= m a

I can get the force of the box when it collides with the spring

F= 12, 425 N

The force the spring makes on the box then is

F = -12,425N = -k d

Then the spring's constant is k= 51,75N/m

To make the two diagrams I need the functions of time when the box slows down

I use the same two equations

V= a t

And

X= v t + 1/2 a t^2

Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.

I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).

Then,

V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec

X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.

for time between 0 and 0,689 sec

Diagrams and equations are in the pictures

7 0
3 years ago
PLEASE HELP ME ASAP PLEASEEEEE
AleksAgata [21]
I believe its the law of inertia
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3 years ago
Can someone help me with the exercise 3 only c,d,e please!!!
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3 years ago
Read 2 more answers
When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam
Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s

b) the change in the combined kinetic energy of the two-car system during this collision

\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))

substitute the value in the equation above

=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J

Hence, the change in combine kinetic energy is -2534.78J

8 0
3 years ago
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