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3 years ago

Find the mass of a football player who weighs 1250 n.

1 answer:

7 0

We know, W = m * g
1250 = m * 9.8
m = 1250 / 9.8
m = 127.55 KG

In short, Your Answer would be 127.55 Kg

Hope this helps!

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A block with mass m = 4.3 kg is attached to two springs with spring constants kleft = 35 N/m and kright = 48 N/m. The block is p

Mrrafil [7]

Answer:

Explanation:

A restoring force is created when a spring is either stretched or compressed.

Restoring force created by first spring = spring constant x contraction

= 35 x .23

= 8.05 N .

It will act towards the right .

Restoring force created by second spring = spring constant x extension

= 48 x .23

= 11.04 N .

It will also act towards the right .

Total force created due to contraction in the first spring and extension in the right spring = 8.05 +11.04 N

= 19.09 N .

4 0

3 years ago

Draw, as best you can, what a velocity graph would look like as a function of time if the acceleration of the object is negative

nlexa [21]

Answer:

Depends on the specific relation of acceleration as function of time, but it would always look like decresing or increasing negatively.

Explanation:

Acceleration is the derivative of velocity with respect to time. That means that it is the change. The fact that is negative means that it is decreasing, or increasing in the negative direction (i.e. going backwards faster).

Attached is the graph with constant negative acceleration, using kinematics relations, it would be a linear equation with negative slope.

5 0

3 years ago

Forces can be added together only if they are

Snezhnost [94]

The answer for question 1 is A.
The answer for question 2 is C

7 0

3 years ago

How far from a -7.80 μC point charge must a 2.40 μC point charge be placed in order for the electric potential energy of the pai

Naddik [55]

Answer:

Distance between two point charges, r = 0.336 meters

Explanation:

Given that,

Charge 1, $q_1=-7.8\ \mu C=-7.8\times 10^{-6}\ C$

Charge 2, $q_2=2.4\ \mu C=2.4\times 10^{-6}\ C$

Electric potential energy, U = -0.5 J

The electric potential energy at a point r is given by :

$U=k\dfrac{q_1q_2}{r}$

$r=k\dfrac{q_1q_2}{U}$

$r=9\times 10^9\times \dfrac{-7.8\times 10^{-6}\times 2.4\times 10^{-6}}{-0.5}$

r = 0.336 meters

So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.

8 0

3 years ago

Determine the values of m and n when the following average distance from the Sun to the Earth is written in scientific notation:

Leviafan [203]

Answer:

$150000000000\ m=1.5\times 10^{11}\ m$

Explanation:

A number can be written in the form of :

$N=m\times 10^n$

Where

m is the real number

n is any integer

In this case, the average distance from the Sun to the Earth is given,

d = 150000000000 meters

There are 10 zeroes in this number. We need to write this number in scientific notation. It is given by :

$d=1.5\times 10^{11}\ m$

So, the average distance from the Sun to the Earth is $d=1.5\times 10^{11}\ m$ . Hence, this is the required solution.

3 0

3 years ago