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2 years ago

Is average speed a scalar or vector quantity? why?

1 answer:

3 0

The average speed is the distance per time ratio (a scalar quantity) Speed is ignorant of direction. Velocity is a vector quantity. It is aware of direction.

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Suppose you were to fill a balloon with air then let go of it withoutntying it closed. What causes the balloon to fly away?

tangare [24]

What causes the ballon to fly away is the air pushing out of the balloon also tied in with air pressure

3 0

2 years ago

Read 2 more answers

A building that is 105.50 ft. feet tall is casting a shadow that is 37.50 ft. feet. If the building next to it, is casting a sha

Arlecino [84]

115.35 ft

Set the proportion up 37.50/105.50 = 41/x and solve for x

7 0

2 years ago

Can you please solve this for me urgently want to make sure if my answers are correct?

MA_775_DIABLO [31]

Answer:

Resolving horizontally. :

$\sum d_{x} = - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\ { \underline{d _{x} = - 5.702 \: m}} \\ \\ \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0 \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y} = - 11.476 \: m}}$

therefore, for resultant:

$d = \sqrt{ {d _{y} }^{2} + d _{x} {}^{2} }$

substitute:

$d = \sqrt{ {( - 5.702)}^{2} + {( - 11.476)}^{2} } \\ \\ d = \sqrt{164.211} \\ \\ { \boxed{ \boxed{ \bf{d = 12.8 \: m}}}}$

6 0

2 years ago

so I know you can solve this either by using Vox or Voy. I'm getting 3.08s when using Vox and 3.14s for Voy way. For Voy I'm usi

Tatiana [17]

The person's horizontal position is given by

$x=v_0\cos40^\circ t$

and the time it takes for him to travel 56.6 m is

$56.6\,\mathrm m=\left(24.0\,\dfrac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t=3.08\,\mathrm s$

so your first computed time is the correct one.

The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity $v_y$ at time $t$ is

$v_y=v_{0y}-gt$

which tells us that he would reach the ground at about $t=3.15\,\mathrm s$ . In this time, he would have traveled

$x=v_{0x}(3.15\,\mathrm s)=57.9\,\mathrm m$

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity $v_{0y}$ would have been a bit smaller than $\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ$ .

7 0

2 years ago

What is the direction of the force for a negative charge moving downward in a magnetic field pointing to the left?

velikii [3]

Answer: A

Out of the screen

Explanation:

Using right hand rule, the magnetic force is perpendicular to the plane form by the magnetic field of a charged particle and its speed. Which will be into the screen.

But the negative charged particle moves in the opposite direction of the positive charged particle. Therefore, the magnetic force direction will be out of the screen

5 0

3 years ago