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Flauer [41]
3 years ago
10

What happens when you fumble out of bounds?

Physics
2 answers:
Mice21 [21]3 years ago
6 0
You give the ball to the other team
IrinaK [193]3 years ago
5 0

in football when you fumble out of bounds the last team that the ball hit off it is the other teams ball

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Which object would be the best insulator of electricity?
solniwko [45]

Answer:

a rubber band

Explanation:

hope this helps!

7 0
2 years ago
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An object has a mass of 6kg. calculate it's gpe​
m_a_m_a [10]

Explanation:

When m=<em>mass</em>

G=<em>a</em><em>c</em><em>c</em><em>e</em><em>l</em><em>e</em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>d</em><em>u</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>gravity</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>H</em><em>=</em><em>h</em><em>e</em><em>i</em><em>g</em><em>h</em><em>t</em>

<em>U</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>f</em><em>o</em><em>r</em><em>m</em><em>u</em><em>l</em><em>a</em>

<em>M</em><em>g</em><em>h</em>

<em>(</em><em>M</em><em>=</em><em>6</em><em>, </em><em>g</em><em>=</em><em>10</em><em>,</em><em>h</em><em>=</em><em>?</em><em>) </em>

6×10×h

=60joules

7 0
2 years ago
The magnitude of the weight of a 3.0 kg object on the surface of the earth is 29 N. True False
madreJ [45]
True

In fact, the weight of an object on the surface of the Earth is given by:
F=mg
where m is the mass of the object and g=9.81 m/s^2 is the gravitational acceleration on Earth's surface. If we use the mass of the object, m=3.0 kg, we find
F=mg=(3.0 kg)(9.81 m/s^2)=29 N
8 0
2 years ago
15 PTS!!!!
Anon25 [30]
I thinks its period because they are on the last column to the right
6 0
3 years ago
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A force of 960 newtons stretches a spring 4 meters. A mass of 60 kilograms is attached to the end of the spring and is initially
Drupady [299]

Answer:

x(t) = - 6 cos 2t

Explanation:

Force of spring = - kx

k= spring constant

x= distance traveled by compressing

But force = mass × acceleration

==> Force = m × d²x/dt²

===> md²x/dt² = -kx

==> md²x/dt² + kx=0   ------------------------(1)

Now Again, by Hook's law

Force = -kx

==> 960=-k × 400

==> -k =960 /4 =240 N/m

ignoring -ve sign k= 240 N/m

Put given data in eq (1)

We get

60d²x/dt² + 240x=0

==> d²x/dt² + 4x=0

General solution for this differential eq is;

x(t) = A cos 2t + B sin 2t   ------------------------(2)

Now initially

position of mass spring

at time = 0 sec

x (0) = 0 m

initial velocity v= = dx/dt=  6m/s

from (2) we have;

dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)

put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

==> 2B = -6

B= -3

Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get

x(t) = - 6 cos 2t

==>  

4 0
3 years ago
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