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ASHA 777 [7]
3 years ago
15

?/1 Jorge traveled 5 miles north to school. He then traveled 3 miles west to the store. Then he left the store and traveled 5 mi

les south to Ted's house. What was Jorge's displacement?
Physics
1 answer:
aleksklad [387]3 years ago
4 0

Answer:

He's 3 miles west of school.

Explanation:

He went 5 miles up and 5 miles down which means that he really didn't go up or down.  In between that, he went 3 miles west so if the 5 milers don't count, this puts him at 3 miles west of school.

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15. Use the chemical equation below to determine how many moles of ammonia
Anastaziya [24]

Answer:

I'm not really sure but I think it is

four

6 0
3 years ago
What happens to myosin and actin as sarcomeres relax?
Leokris [45]
The myosin heads pull on the actin, bringing them closer together

4 0
3 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
2 years ago
A 5.00 kilogram block slides along a horizontal,frictionless surface at 10.0 meters per second. for 4.00 seconds. The magnitude
finlep [7]
In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
                   = (5 * 10) Kg m/s
                   = 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".
5 0
3 years ago
Consider a sphere and a cylinder of equal volume made of copper. Both the sphere and the cylinder are initially at the same temp
zepelin [54]

Answer: The cylinder

Explanation:

Among all other solid shapes, the sphere has the smallest area for a given volume.

By experiment, the ratio of the radius of a sphere to a cylinder of equal volume is less than 1.

Recall;

That the Rate of transfer of convective heat (Q) = h × A ×change in temperature.

Where ,

h= the co efficient of convective heat transfer

A= the cross sectional area.

As such, since the sphere has a smaller surface area relative to the cylinder, the sphere transfers heat slower than the cylinder.

Therefore, if the sphere and cylinder are exposed to convection in the same environment, then, the cylinder cools faster.

PS; the more the Area, the higher the rate of heat transfer and vice versa.

7 0
3 years ago
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