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xxTIMURxx [149]
3 years ago
10

What are two types of scientific investigations?

Physics
1 answer:
kvv77 [185]3 years ago
4 0
Experiments and fieldwork
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a man stands in a lift going downward with uniform velocity. he experiences a loss of weight at the start but not when lift is i
AnnyKZ [126]

Answer:

It is explained in the explanation section

Explanation:

When the lift starts going downwards, it will start accelerating downwards. After a while, it will start moving with a constant velocity.

Constant velocity means that acceleration is zero and so the man will not feel any weight loss.

Now, Once the lift achieves constant velocity the acceleration is zero hence he will not experience any weight loss.

However, when the lift is in uniform motion, the lift and the man will fall down with an acceleration(a) that is less than that due to gravity(g) . Thus, the man will feel an apparent weight F which is not equal to zero.

6 0
3 years ago
Which statement is the most appropriate conclusion for David to write?
ycow [4]
I need more to answer this
4 0
2 years ago
A 0.49-kg cord is stretched between two supports,7.3m apart. When one support is struck by a hammer, a transverse wave travels d
Wewaii [24]

Answer:

T= 5.18N

Explanation:

u = mass of chord / length of chord

u = 0.49/ 7.3

u = 0.067 kg/m

Velocity of sound waves (v) =length of chord / time taken for wave to travel

v = 7.3 / 0.83 = 8.795m/s

Tension is calculated below using the formula

T = v² * u

T = (8.795)² x 0.067

T= 5.18N

3 0
3 years ago
Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10
gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

\Sigma \tau=0

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

-\tau_l+\tau_p+\tau_{cm}=0          (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

\tau=Fd           (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N

The force applied by the left support is 2123.33 N

8 0
3 years ago
You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. Th
Rama09 [41]

Answer:

 I = 8.75 kg m

Explanation:

This is a rotational movement exercise, let's start with kinetic energy

        K = ½ I w²

They tell us that K = 330 J, let's find the angular velocity with kinematics

      w² = w₀² + 2 α θ

as part of rest w₀ = 0

      w = √ 2α θ

let's reduce the revolutions to the SI system

      θ = 30.0 rev (2π rad / 1 rev) = 60π rad

let's calculate the angular velocity

      w = √(2  0.200  60π)

      w = 8.683 rad / s

we clear from the first equation

        I = 2K / w²

let's calculate

        I = 2 330 / 8,683²

        I = 8.75 kg m

4 0
2 years ago
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