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Pani-rosa [81]
3 years ago
15

The purple arrow shows speed and direction of the ball. This arrow is called what?

Physics
2 answers:
valina [46]3 years ago
6 0
Vector I'm pretty certain
Lapatulllka [165]3 years ago
5 0

The arrow is called vector

You might be interested in
The middle one please need done in 45min
netineya [11]

Answer:

600 J

Explanation:

it's obviously btw so yeahhhh

5 0
3 years ago
A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a ra
worty [1.4K]

Answer:

The torque on the pulley, when the system is motionless is approximately 9.81 N·m

Explanation:

The given parameters are;

The mass of the object = 2 kg

The friction between the rope and the pulley = 0

The mass of the rope, m_r = 0.5 kg

The mass of the pulley, m_p = 0.01 kg

The radius of the pulley, r = 0.25 m

The torque on the pulley, τ = I·α = F × D

The torque on the pulley, when the system is motionless, τ = F × D

Where;

F = The force acting on the pulley rope = The weight of the mass ≈ 2 kg × 9.81 m/s² = 19.62 N

D = The diameter of the pulley = 2×r = 2 × 0.25 m = 0.5 m

Therefore;

τ = 19.62 N × 0.5 m = 9.81 N·m

The torque on the pulley, when the system is motionless, τ ≈ 9.81 N·m.

4 0
3 years ago
Dolphins communicate using compression waves (longitudinal waves). Some of the sounds dolphins make are outside the range of hum
lana66690 [7]

Answer: send the message underwater because a more dense medium would make the sound travel faster.

Explanation:

Dolphins communicate using compression waves - longitudinal waves. Longitudinal waves requires a medium to travel. A longitudinal wave transfers energy by the vibration of medium particles in the direction of the wave motion. Compression are the regions where density of the medium is higher and rarefaction is a low density region.

A longitudinal wave travels faster in a denser medium. It has maximum speed in solid and minimum in gas. Thus, to transfer message quickly to dolphin B., dolphin A should send the message underwater and not in air. This is because water has higher density than air. Molecules collide more quickly in water than in air and it takes less time for signal to travel.

3 0
3 years ago
Jupiter’s Great Red Spot rotates completely every six days. If the spot is circular (not quite true, but a reasonable approximat
artcher [175]

Answer:

v = 567.2 km/h

Explanation:

As we know that if Jupiter Rotate one complete rotation then it means that the it will turn by 360 degree angle

so here the distance covered by the point on its surface in one complete rotation is given by

distance = 2\pi r

distance = \pi D

now we will have the time to complete the rotation given as

t = 6 days

t = 6 (24 h) = 144 h

now the speed is given by

speed = \frac{distance}{time}

speed = \frac{\pi D}{t}

speed = \frac{\pi(26000 km)}{144}

v = 567.2 km/h

5 0
3 years ago
As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval t=14.86\times10^{-3}\ s

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

9.8\times1.94=\dfrac{1}{2}\times v^2

v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

v'=5.38\ m/s

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

Hence,The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

6 0
3 years ago
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