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wolverine [178]
3 years ago
10

A horizontal spring with spring constant 950 N/m is attached to a wall. An athlete presses against the free end of the spring, c

ompressing it 7.0 cm. How hard is the athlete pushing?
Physics
1 answer:
OLga [1]3 years ago
4 0

Answer:

66.5N

Explanation:

F = kx

Where F = force

K = spring constant

x = compression

Given

K = 950N/m

x = 7.0cm

F = ?

First convert the compression to meters .

7.0cm = 7.0 x 0.01

= 0.07 meters

Therefore

F = 950 x 0.07

= 66.5N

You might be interested in
A cube of wood having an edge dimension of 20.0cm and a density of 650 kg /m³ floats on water. (b) What mass of lead should be p
schepotkina [342]

The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.

<h3>What mass of lead should be placed on the cube?</h3>

Given: Side of the cube (a) = 20cm

The density of the cube (ρc) = $$650 kg/m^3

a) Applying the force balance, the buoyant force must be equal to the weight of the cube

ρcgV = ρg × (Ax)

Substituting the values in the above equation, we get

(650*(0.2 m)^3)=1000*(0.2 m)^2*x

x = 0.13

where x is the height of the cube in the water

$$A = a^2 is the area of the cross-section

ρ is the density of the water

V is the volume of the cube

Now, the height above the surface of the water would be

h = a − x

Substituting the values, then we get

h = 0.2 − 0.13

h = 0.07 m

b) The mass added is "m" so the complete cube is submerged in the water, therefore

ρcgV + mg = ρg × (V)

$(650*(0.2 m)^3)+m=1000*(0.2 m)^3

m = 2.8 kg

The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.

To learn more about buoyant force refer to:

brainly.com/question/11884584

#SPJ4

7 0
1 year ago
A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto
Ilya [14]
Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer: 
0.32 m (single amplitude), or
0.64 m (double amplitude)

6 0
2 years ago
A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .
zimovet [89]

Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

F=m\times a

F=m\times (\dfrac{v-u}{t})

F=0.003\times (\dfrac{30}{0.06})

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

6 0
3 years ago
Why is your State have dimensions ​
Ilya [14]

Answer:

not sure you

Explanation:

8 0
3 years ago
We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,
larisa [96]

Answer:

The final charges of each sphere are:   q_A = 3/8 Q , q_B = 3/8 Q ,               q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

                q_A = Q / 2

                q_B = Q / 2

Now sphere A touches sphere C, ending with half the charge

                q_A = ½ (Q / 2) = ¼ Q

                q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

                  q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

                  q_A = 3/8 Q

                  q_B = 3/8 Q

The final charges of each sphere are:

                q_A = 3/8 Q

                q_B = 3/8 Q

                q_C = 3/4 Q

7 0
3 years ago
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