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____ can be calculated if you know the distance that an object travels in one unit of time. A.motion
B.meter
C.Rate
D.Speed
E.velocity
F.slope
G.refrence point

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____ can be calculated if you know the distance that - 1

Answers

Answer 1
Answer:

Speed can be calculated if you know the distance that an object travels in one unit of time, therefore the correct answer is option D.

What is speed?

The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object.

The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.

Thus, Speed can be calculated if you know the distance that an object travels in one unit of time, therefore the correct answer is option D.

Learn more about speed from here, refer to the link;

brainly.com/question/7359669

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Answer 2
Answer:

Answer:

D.Speed

Explanation:

The speed of an object is the distance the object travels in one unit of time.

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A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?
Ball 1 is launched with an initial vertical velocity v1 = 146 ft/sec. Ball 2 is launched 2.3 seconds later with an initial vertical velocity v2. Determine v2 if the balls are to collide at an altitude of 234 ft. At the instant of collision, is ball 1 ascending or descending?
A steel cable lying flat on the floor drags a 20 kg block across a horizontal, frictionless floor. A 100 N force applied to the cable causes the block to reach a speed of 4.2 m/s in a distance of 2.0 m.What is the mass of the cable?
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The planet earth orbits around the sun and also spins around its own axis. 33% part (a) calculate the angular momentum of the earth in its orbit around the sun in kg • m2/s

a trcuk weighs four times as much as a stationary car. if teh truck coasts into the car at 12 km/s and they stick toegther, what is their final velocity

Answers

Answer:

  v=9.6 km/s

Explanation:

Given that

The mass of the car =  m

The mass of the truck = 4 m

The velocity of the truck ,u= 12 km/s

The final velocity when they stick = v

If there is no any external force on the system  then the total linear momentum of the system will be conserve.

Pi = Pf

m x 0 + 4 m x 12 = (m + 4 m) x v

0 + 48 m = 5 m v

5  v  = 48

v=(48)/(5)\ km/s

v=9.6 km/s

Therefore the final velocity will be 9.6 km/s.

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a mass of m2 = 91 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is µs = 0.79 and the coefficient of kinetic friction between the two crates is µk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s25)As the upper crate slides, what is the acceleration of the lower crate?

Answers

Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

=  μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

The digital exchange of structured data is called ?

Answers

Answer:

Electronic data interchange

What is the motion of the particles in this kind of wave? A hand holds the left end of a set of waves. The waves themselves make a larger set of waves in the same direction as that of the smaller waves. A label Wave motion is above the series of waves and an arrow next to the label points right. The particles will move up and down over large areas. The particles will move up and down over small areas. The particles will move side to side over small areas. The particles will move side to side over large areas.

Answers

Answer:

→A←

Explanation:

D its incorrect in edge

Answer:

D

Explanation:

The particles will move side to side over large areas

Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculate the force of gravity that the space shuttle experiences

Answers

Final answer:

The force of gravity that the space shuttle experiences is 9.8 x 10^5 Newtons.

Explanation:

To calculate the force of gravity that the space shuttle experiences, we can use the equation F = mg, where F represents the force of gravity, m is the mass of the object, and g is the acceleration due to gravity (approximately 9.8 m/s² on Earth). In this case, the mass of the space shuttle is given as 1.0 x 10^5 kg. However, we need to convert the altitude of the shuttle into meters, so 200.0 km becomes 200,000 meters.

Now we can calculate the force of gravity:

F = (1.0 x 10^5 kg)(9.8 m/s²)

F = 9.8 x 10^5 N

Therefore, the space shuttle experiences a force of gravity of 9.8 x 10^5 Newtons.

Learn more about force of gravity here:

brainly.com/question/12753714

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A squirrel is running a race where she is on track for her average velocity to be 6.0 m/s. She is distracted by a dummy of an attractive male squirrel and pauses for 3.0 s. As a result her average velocity ends up being 5.0 m/s instead. What is the length of the race? [HINT: construct two equations with the same two unknowns in them and you can solve the system of equations]

Answers

Answer:

90 m

Explanation:

There are two unknowns: the amount of time the squirrel spent running, and the length of the race.  Let's call these t and x, respectively.

The average velocity is the total distance divided by the total time.

5.0 m/s = x / (t + 3.0)

The total distance is the time she spent running times the speed she ran at.

x = (6.0 m/s) t

Substitute and solve:

5 = 6t / (t + 3)

5 (t + 3) = 6t

5t + 15 = 6t

t = 15

She ran for 15 seconds (not including the 3 seconds she stopped).  So the length of the race is:

x = (6.0 m/s) (15 s)

x = 90 m
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