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2 years ago

The net vertical force on a box F as a function of the vertical position y is shown below.

Physics

1 answer:

Ray Of Light [21]2 years ago

4 0

Answer:

W = 0 J

Explanation:

Formula for work done is;

W = F × d

Where;

W is work done

F is Force

d is distance covered

What this means is that we will calculate the area under which the given times in the graph pass.

Thus;

At constant force of F = 40 N which falls in between distance of 0 m and 2m,

W1 = 40 × 2 = 80 J

At constant force of -20 n which will fall between distance of 2 m and 6 m which is 4m, we have;

W2 = -20 × 4

W2 = -80 J

Thus, total workdone is;

W = W1 + W2

W = 80 - 80

W = 0 J

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A torque of 0.77 N⋅m is applied to a bicycle wheel of radius 30 cm and mass 0.70 kg

Naddik [55]

Answer:

α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²

Explanation:

4 0

3 years ago

A net force of 20 N acting on a wooden block produces an

Ymorist [56]

Answer:

From the second law of motion:

F = ma

we are given that the force applied on the block is 20N and the block accelerates at an acceleration of 4 m/s/s

So, F= 20N and a = 4 m/s/s

Replacing the variables in the equation:

20 = 4* m

m = 20 / 4

m = 5 kg

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3 years ago

You are riding in an airplane that is moving at a speed of 600 km/h. If you walk from the FRONT of the plane TOWARD the BACK of

MatroZZZ [7]

Answer:

600km/h as u are on a platform moving at the speed of 600 km/h where u are moving in relativity to the plane it's self.

5 0

3 years ago

A 0.20 kg mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured

xxMikexx [17]

Answer:

b. 0.20 m/s.

Explanation:

Given;

initial mass, m = 0.2 kg

maximum speed, v = 0.3 m/s

The total energy of the spring at the given maximum speed is calculated as;

K.E = ¹/₂mv²

K.E = 0.5 x 0.2 x 0.3²

K.E = 0.009 J

If the mass is changed to 0.4 kg

¹/₂mv² = K.E

mv² = 2K.E

$v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s$

Therefore, the maximum speed is 0.20 m/s

4 0

3 years ago

Need help on physics/momentum

koban [17]

Answer:

vf=11.67m/s

Explanation:

vf:

Δp=m(vf-vi)

80=12(vf-5)

80=12vf-60

140=12vf

vf=11.67m/s

pf:

p=m*v

p=12*11.67

p=140.04 N*s

Hope this helps

5 0

3 years ago