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Natalka [10]
3 years ago
7

A baby carriage is rolling down a hill. Take into consideration, the carriage has 90J of kinetic energy and the mass of the carr

iage is 0.5m. If you can only run at a speed of 10m/s, will you catch the baby carriage? What is the velocity of the carriage?
Physics
1 answer:
Sphinxa [80]3 years ago
7 0

Answer:

YES. 6√10

Explanation:

apply E=1/2mv^2 and find the velocity of carriage which is 6√10 ms-1

as our velocity is higher than that of cart. we can catch the carriage

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If a dog is running at 4.38 m/s for 37.34 seconds, how far will it go?
sdas [7]

Answer:

163.5m

this is how far the doggo would go

(In theory)

If I'm wrong I apologize

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3 years ago
What is nuclear energy​
suter [353]

Answer:

Nuclear energy, also called atomic energy, energy that is released in significant amounts in processes that affect atomic nuclei, the dense cores of atoms. It is distinct from the energy of other atomic phenomena such as ordinary chemical reactions, which involve only the orbital electrons of atoms.

Explanation:

7 0
3 years ago
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The odometer of a car changes from 1048 km to 1096 km in 40
Makovka662 [10]

Answer:

20m/s

Explanation:

it covers 20 metres in a second

3 0
2 years ago
A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o
marshall27 [118]

Answer:

The magnitude of the acceleration of the box is 2 m/s².

Explanation:

Given:

Mass of the box, m=5.0 kg

Force acting towards east, F=27 N

Frictional force acting towards west, f=17 N

Let the acceleration be a m/s².

Now, net force acting on the box towards east is given as:

F_{net}=F-f=27-17=10\textrm{ N}

From Newton's second law of motion,

F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2

Therefore, the magnitude of the acceleration of the box is 2 m/s².

6 0
3 years ago
Read 2 more answers
The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca
zlopas [31]

Answer:

The value is  I = 0.0932 ML ^2  

Explanation:

From the question we are told that

  The rotational inertia about one end is I_R =  \frac{1}{3} ML^2

   The location of the axis of rotation considered is d =  0.4 L

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is  0.4 M

Generally the length of the rod from the its beginning to the axis of rotation consider is

      k = 1 - 0.4 L = 0.6L

Generally the mass of the portion  of the rod from the its beginning to the axis of rotation consider is

    m  =  1- 0.4 M = 0.6 M

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

     I_{R1} =  \frac{1}{3} (0.6 M )(0.6L)^2

    I_{R1} =  \frac{1}{3} (0.6 M )L^2 0.6^2

Generally the rotational inertia about the axis of rotation consider for the second  portion of the rod is

     I_{R2} =  \frac{1}{3} (0.6 M )(0.6L)^2

=> I_{R2} =  \frac{1}{3} (0.4 M )(0.4L)^2

=>  I_{R2} =  \frac{1}{3} (0.4 M )L^2 0.4^2

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

  I =   \frac{1}{3} (0.6 M )L^2 0.6^2 +   \frac{1}{3} (0.4 M )L^2 0.4^2

=>   I =  \frac{1}{3} ML ^2  [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]

=>   I = 0.0932 ML ^2  

8 0
3 years ago
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