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2 years ago

When might it be harder to stop a vehicle moving at 30 km/h than one moving at 60 km/h?

Physics

1 answer:

rodikova [14]2 years ago

4 0

Answer:

when the momentum of the vehicle moving at 30 km/h is higher than the one from the vehicle moving at 60 km/h

Explanation:

It's much harder to stop a freight truck moving at 30 km/h than a hot wheels car moving at 60 km/h.

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A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert

pychu [463]

(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

$\theta=65.4^{\circ}$

The projection of the length of the pendulum along the vertical direction is

$L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm$

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

$\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m$

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

$\Delta U = mg \Delta y$

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

$\Delta y = 0.211 m$ is the vertical displacement of the pendulum

So, the work done by gravity is

$W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J$

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

$\theta=90^{\circ}, cos \theta = 0$

and so the work done is zero.

5 0

3 years ago

The mass number is determined by which particles?

BartSMP [9]

I am not completely sure, but I believe that it depends on the total mass of the Protons and Neutrons

7 0

3 years ago

Read 2 more answers

What would happen if the pilot did not keep the airplane "trimmed"

Mademuasel [1]

Answer:

In explanation

Explanation:

Pilots who dont use trim often like the feeling of holding constant back pressure because The heavier control forces makes it more difficult to over-control the airplane inside the turn, so it gives the sense of a more stable flight

3 0

3 years ago

True or false Scientific endeavor is driven by both simple curiosity as well as societal demands

ICE Princess25 [194]

Answer:

its true that Scientific endeavor is driven by both simple curiosity as well as societal demands.

Explanation:

When a scientist has a curiosity about something he carried out a research. and when their is a demand of something in society that time scientific research is carried out. Therefore its true that a scientific endeavor is driven by simple curiosity or societal demand.

For example

in society, there is demand of a medicine which can completely kill the cancer and a scientist has curiosity to know how to kill cancer cell. In this way a scientific endeavor for cancer medicine can be carried out by both simple curiosity as well as societal demands.

6 0

3 years ago

You're driving your new sports car at 80 mph over the top of a hill that has a radius of curvature of 540 m. What fraction of yo

luda_lava [24]

Answer:

75.84%

Explanation:

We were given Speed of the sports car, v as 80 mph , we can convert to m/s for unit consistency.

v=80mph= 35.76 m/s

The radius of curvature is given as , r = 540 m

✓ the normal weight can be denoted as Wn

✓ the apparent weight of the person can be denoted as Wa

Wn= normal weight= mg

Wa=apparent weight = (mg - mv^2/r)

g= acceleration due to gravity= 9.8m/s^2

The apparent weightand normal weight has a ratio of

Mn/Ma= [mg - mv^2/r]/mg ........eqn(1)

If we simplify eqn(1) we have

Mn/Ma=[g - vr^2/g].............eqn(2)

Then substitute the given values

Mn/Ma=9.8 - [(35.76^2)/540]/ 9.8

=0.758×100%

Mn/Ma=75.84%

Hence, the required fraction is 75.84%

4 0

2 years ago