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2 years ago

Which wave must have a medium to travel?

Physics

2 answers:

o-na [289]2 years ago

6 0

Letter B

without a medium, there is nothing to compress, hence, no wave. A fast- medium like a gas (air) is easy to compress and allows waves to move through it easily. a slow medium, like a liquid, is still pretty fast, but not as fast as air.

sladkih [1.3K]2 years ago

3 0

The answer is A good luck !

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An object has a kinetic energy of 175 J and a momentum of magnitude 25.0 kg m/s. Find the

DedPeter [7]

Answer:14 m/s

Explanation:

Kinetic energy(ke)=175J

Momentum(M)=25kgm/s

Speed=v

Mass=m

Ke=(m x v x v)/2

175=(mv^2)/2

Cross multiply

175 x 2=mv^2

350=mv^2

Momentum=mass x velocity

25=mv

m=25/v

Substitute m=25/v in 350=mv^2

350=25/v x v^2

350=25v^2/v

v^2/v=v

350=25v

v=350/25

v=14 m/s

5 0

3 years ago

2. A stone is thrown vertically upward with a speed of 22m/s.

Eduardwww [97]

Answer:

Explanation:

Energy E is conserved:

$E=\frac{1}{2}mv^2+mgh$

If v₀ = 22m/s, h₀=0m and h₁=25m:

$E=\frac{1}{2}mv_0^2=\frac{1}{2}mv_1^2+mgh_1$

Solving for v₁:

$v_1=\sqrt{v_0^2-2gh_1}$

There is no real solution, because the stone never reaches 25m.

7 0

3 years ago

A car of mass 600kg is moving at 15m/s. The driver accelerate gently to a final velocity of 30m/s so that a force of force acts

Anvisha [2.4K]

Answer:

Really hope you get it!!

3 0

3 years ago

Read 2 more answers

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of 4.15 m from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height 4.15m from the ground before dropping down.

3 0

2 years ago

A truck is moving around a circular curve at a uniform velocity of 13 m/s. If the centripetal force on the truck is 3,300 N and

Paladinen [302]

Answer: Option B.

Since here the truck is moving on a circular track, it will experience centripetal force.

F(centripetal) = m × acc
or
$r = \frac{m v^{2}}{F}$

where r is the radius of the track.
m is the mass of truck
v is the speed of the truck.
Given: v = 13 m/s
m = 1,600 kg
F = 3300 Newton

To find = radius of track=?
$r = \frac{m v^{2} }{F}$
$r = \frac{1600*13*13}{3300}$
r = 81.94 m
Therefore, radius of track is 81.94 m

4 0

3 years ago

Read 2 more answers