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VladimirAG [237]
3 years ago
15

If 1 foot is 30.28 centimeters.how many cm is 130 feet

Physics
1 answer:
uranmaximum [27]3 years ago
6 0

Answer:3,936.4

Explanation:

If 1 foot is 30.28 centimeters multiply 30.28 by 130

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13. An object with a mass of 2.0 kg has a force of 4.0 newtons applied to it.
julsineya [31]

Answer:2m/s²

Explanation: Well F=MA so sice F=4N and M=2kg let's plug in the values

4N=2KG*A

A=4N/2KG

A=2m/s²

7 0
3 years ago
A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo le
Alona [7]

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

R = u² sin2θ/g

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

u² = \frac{Rg}{sin2\theta}

u^2 = \frac{9.1 x 9.80}{sin26}

u^2 = 113.17

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

v^2 = (v_osin\theta)^2 -2gh

h =\frac{(v_o^2sin\theta)^2}{2g}

h  =  \frac{(113.17)(sin26)^2}{(2 x 9.80)}}

h = 1.10 m

7 0
3 years ago
A force of 8.0 N is along x direction, another force of 6.0 N is along +y direction. If both forces are acting on a point object
Darya [45]

Answer:

Resultant force, R = 10 N

Explanation:

It is given that,

Force acting along +x direction, F_x=8\ N

Force acting along +y direction, F_y=6\ N

Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,

R=\sqrt{F_x^2+F_y^2+F_xF_y\ cos\theta}

Here \theta=90^{\circ}

R=\sqrt{F_x^2+F_y^2}

R=\sqrt{8^2+6^2}

R = 10 N

So, the resultant force on the object is 10 N. Hence, this is the required solution.

6 0
3 years ago
The coach has not seen improvement as expected in his team and plans to change practice session. To help the players he changes
Allushta [10]

Answer:

progession

Explanation:

7 0
3 years ago
A rock is thrown straight down, not dropped, from the roof of a building that is 61 m above the ground. If it takes 3.1 s to rea
PtichkaEL [24]
Hiii !!!
I am sending the soluction !!!

If you have any question, let me now =)

Jorge:)

4 0
3 years ago
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