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3 years ago

A dog runs out the back door of house and stops and grabs a bone.Then the dog returns

Physics

2 answers:

Katen [24]3 years ago

7 0

B) 4 seconds because the position did not change between 4 and 8 seconds

Helga [31]3 years ago

4 0

i would say 4 seconds because according to the graph the flat line is where the dog stopped.

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A 20 kilogram ball rolls down a 10 meter ramp at the rate of 15 meters per second. The kinetic energy is joules

kobusy [5.1K]

Answer:2250J

Explanation:

mass(m)=20kg

velocity(v)=15m/s

Kinetic energy=(m x v^2)/2

Kinetic energy =(20 x 15^2)/2

Kinetic energy =(20x15x15)/2

Kinetic energy=4500/2

Kinetic energy=2250J

3 0

4 years ago

Considering the various theories, the energy used in forming organic molecules in the primitive atmosphere could have come from

OLEGan [10]

Answer:

<h2>e. sound. </h2>

Explanation:

Such type of atmosphere in which oxygen was not present or was present in little amount is known as the primitive atmosphere and such type of atmosphere was present in the initial stage of the earth formation.
During this period, water vapor, nitrogen, hydrogen and carbon dioxide gases were present.
These gases interact with the help of energy that comes from many sources such as lightning, ultraviolet radiation, electric spark and some other.
When these inorganic molecules react in the then the formation of organic compounds takes place that becomes the basis of the organization of the life on the earth and called an organic evolution of life.

8 0

3 years ago

An object with an initial velocity of 10 m/s accelerates at a rate of 3.5 m/s2 for 8 seconds. How far will it have traveled duri

ICE Princess25 [194]

Answer:

Distance, d = 192 meters

Explanation:

We have,

Initial velocity of an object is 10 m/s

Acceleration of the object is 3.5 m/s²

Time, t = 8 s

We need to find the distance travelled by the object during that time. Second equation of motion gives the distance travelled by the object. It is given by :

$d=ut+\dfrac{1}{2}at^2$

$d=10\times 8+\dfrac{1}{2}\times 3.5\times 8^2\\\\d=192\ m$

So, the distance travelled by the object is 192 meters.

4 0

3 years ago

A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0

3 years ago

A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed

Tresset [83]

For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
-24.96 = (-0.65) M2
M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg

4 0

3 years ago