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3 years ago

1. Transverse waves _____.

Physics

2 answers:

poizon [28]3 years ago

8 0

D. have particle moment perpendicular to the direction of the energy.

kumpel [21]3 years ago

7 0

The answer is D to you question

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The Fitness Gram push-up test is a measure of

Morgarella [4.7K]

Answer:

The answer is B.

Explanation:

I meant B. not C so sorry

6 0

3 years ago

A 2200-kg railway freight car coasts at 4.1 m/s underneath a grain terminal, which dumps grain directly down into the freight ca

777dan777 [17]

Answer:

The answer is " $2.41 \times 10^3$ "

Explanation:

Given:

$m_i = 2000 \ kg \\\\v_i= 4.1 \ \frac{m}{s} \\\\v_f = 3.4 \ \frac{m}{s} \\$

Using formula:

$\to m_iv_i = m_fv_f \\\\\to m_f= \frac{m_iv_i}{v_f}$

$p_i, p_f$ = system initial and final linear momentum.

$V_i, v_f$ = system original and final linear pace.

$m_i$ = original weight of the car freight.

$m_f$ = car's maximum weight

$= \frac{ 2000 \times 4.1}{3.4}\\\\= \frac{ 8.2\times 10^3}{3.4}\\\\= 2.41 \times 10^3$

$\boxed{m_f = 2.41 \times 10^3}$

8 0

2 years ago

If a ????=87.5 kgm=87.5 kg person were traveling at ????=0.900????v=0.900c , where ????c is the speed of light, what would be th

Diano4ka-milaya [45]

Answer:

$\frac{K.E_r}{K.E}=2.875$

Explanation:

Given:

mass, m = 87.5kg

Velocity, V = 0.900c

now,

the relativistic kinetic energy id given as:

$K.E_r=(\gamma-1)mc^2$ ...........(1)

where,

$\gamma$ = relativistic factor, given as; $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Now, the classical kinetic energy is given as:

$K.E = \frac{1}{2}mv^2$ ..........(2)

Dividing the equation (1) by (2) we get

$\frac{K.E_r}{K.E}=\frac{(\gamma-1)mc^2}{\frac{1}{2}mv^2}$

$\frac{K.E_r}{K.E}=\frac{(\gamma-1)c^2}{\frac{1}{2}v^2}$

substituting the values in the equation we get,

$\frac{K.E_r}{K.E}=\frac{(\frac{1}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}-1)c^2}{\frac{1}{2}\times(0.90c)^2}$

$\frac{K.E_r}{K.E}=2.875$

5 0

3 years ago

While looking at bromine (Br) on the periodic table, a student needs to find another element with very similar chemical properti

julsineya [31]

Answer: 3) There are many possible elements, and they are all in the same vertical column as bromine.

Explanation:

4 0

3 years ago

Read 2 more answers

Starting from rest near the surface of the Earth, a 25-kg beam slides 12 m down a vertical pine tree, and has a speed of 6 m/s j

Mamont248 [21]

Answer:

The frictional force acting on the bear during the slide is 207.5 N

Explanation:

Given;

mass of beam, m = 25-kg

vertical height, h = 12 m

speed of fall, v = 6 m/s

Change in potential energy of the beam:

ΔP.E = -mgh = - 25 x 9.8 x 12 = -2940 J

Change in kinetic energy of the beam:

Δ K.E = ¹/₂mv² = ¹/₂ x 25 x (6)² = 450 J

Change in thermal energy of the system due to friction:

ΔE = - (ΔP.E + Δ K.E)

ΔE = - (-2940 J + 450 J)

ΔE = 2940 J - 450 J = 2490 J

Frictional force (in N) acting on the bear during the slide:

F x d = Fk x h = ΔE

Where;

Fk is the frictional force

Fk = ΔE/h

Fk = 2490J / 12m

Fk = 207.5 N

Therefore, the frictional force acting on the bear during the slide is 207.5 N

3 0

3 years ago