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Irina18 [472]
2 years ago
15

If the northern hemisphere were tilted 90° towards the sun which location would be warmer in summer the Arctic Circle or Florida

Physics
1 answer:
nikdorinn [45]2 years ago
7 0
Florida would be warmer.
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Juan was wearing a bright red shirt in a very dark room. What color did his shirt appear to the people with him in the room? A)
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It would appear black.   

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What major region does this profile<br> most likely represent?
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A mid ocean ridge possibly a plate margin spreading area

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A spring is used as part of a lift system and follows Hooke's law. If the spring is
salantis [7]

Answer:

1.05m or 105cm

Explanation:

Using the hooke's law equation as follows;

F = –k.x

Where;

F = force (N)

x = extension length (m)

k = constant of proportionality (N/m)

According to the information given in this question;

Displacement (x) = 85cm = 85/100 = 0.85m

Force = 12500N

Using F = kx, we find the proportionality constant

k = F/x

K = 12500/0.85

K = 14705.8N/m.

Also, since K = 14705.8N/m, the displacement (x), when the force increases to 15500N is;

F = kx

x = F/k

x = 15500/14705.8

x = 1.05m or 105cm

6 0
2 years ago
The parking brake on a 1000 kg Cadillac has failed, and it is rolling slowly, at 1 mph , toward a group of small children. Seein
iogann1982 [59]

Answer:

0.5 mph in the opposite direction

Explanation:

m_1 = Mass of Cadillac = 1000 kg

v_1 = Velocity of Cadillac = 1 mph

m_2 = Mass of Volkswagen = 2000 kg

v_2 = Velocity of Volkswagen

In order to know the speed the system must have the momentum exchange

As the linear momentum of the system is conserved

m_1v_1+m_2v_2=0\\\Rightarrow v_2=-\dfrac{m_1v_1}{m_2}\\\Rightarrow v_2=\dfrac{1000\times 1}{2000}\\\Rightarrow v_2=-0.5\ mph

The speed of the impact is given by 0.5 mph in the opposite direction

8 0
3 years ago
A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.650 kg ,
Maru [420]

Answer:

a) Coefficient of kinetic friction between block and surface = 0.12

b) Decrease in kinetic energy of the bullet = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = 0.541 J

Explanation:

Given,

Mass of bullet = 4.00 g = 0.004 kg

Initial velocity of the bullet = 400 m/s

Mass of wooden block = 0.65 kg

Initial velocity of the wooden block = 0 m/s (since it was initially at rest)

Final velocity of the bullet = 190 m/s

Distance slid through by the block after the collision = d = 72.0 cm = 0.72 m

Let the velocity of the wooden block after collision be v

According to the law of conservation of momentum,

Momentum before collision = Momentum after collision

Momentum before collision = (Momentum of bullet before collision) + (Momentum of wooden block before collision)

Momentum of bullet before collision = (0.004×400) = 1.6 kgm/s

Momentum of wooden block before collision = (0.65)(0) = 0 kgm/s

Momentum after collision = (Momentum of bullet after collision) + (Momentum of wooden block after collision)

Momentum of bullet after collision = (0.004×190) = 0.76 kgm/s

Momentum of wooden block after collision = (0.65)(v) = (0.65v) kgm/s

Momentum balance gives

1.6 + 0 = 0.76 + 0.65v

0.65v = 1.6 - 0.76 = 0.84

v = (0.84/0.65)

v = 1.29 m/s

The velocity of the wooden block after collision = 1.29 m/s

To obtain the coefficient of kinetic friction between block and surface, we will apply the work-energy theorem.

The work-energy theorem states that the work done in moving the block from one point to another is equal to the change in kinetic energy of the block between these two points.

The points to consider are the point when the block starts moving (immediately after collision) and when it stops as a result of frictional force.

Mathematically,

W = ΔK.E

W = workdone by the frictional force in stopping the wooden block (since there is no other horizontal force acting on the block)

W = -F.d (minus sign because the frictional force opposes motion)

d = Distance slid through by the block after the collision = 0.72 m

F = Frictional force = μN

where N = normal reaction of the surface on the wooden block and it is equal to the weight of the block.

N = W = mg

F = μmg

W = - μmg × d = (-μ)(0.65)(9.8) × 0.72 = (-4.59μ) J

ΔK.E = (final kinetic energy of the block) - (initial kinetic energy of the block)

Final kinetic energy of the block = 0 J (since the block comes to a rest)

(Initial kinetic energy of the block) = (1/2)(0.65)(1.29²) = 0.541 J

ΔK.E = 0 - 0.541 = - 0.541 J

W = ΔK.E

-4.59μ = -0.541

μ = (0.541/4.59)

μ = 0.12

b) The decrease in kinetic energy of the bullet

(Decrease in kinetic energy of the bullet) = (Kinetic energy of the bullet before collision) - (Kinetic energy of the bullet after collision)

Kinetic energy of the bullet before collision = (1/2)(0.004)(400²) = 320 J

Kinetic energy of the bullet after collision = (1/2)(0.004)(190²) = 72.2 J

Decrease in kinetic energy of the bullet = 320 - 72.2 = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = (1/2)(0.65)(1.29²) = 0.541 J

Hope this Helps!!!

4 0
2 years ago
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