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umka2103 [35]
3 years ago
13

The graph shows the amplitude of a pausing wave over time in secondo (o).

Physics
2 answers:
kipiarov [429]3 years ago
4 0

Answer:

The graph shows the amplitude of a pausing wave over time in secondo (o).

What is the approximate frequency of the wave shown?

A. 0.5 Hz

B. 0.1 Hz

C. 9 Hz

D. 20Hz

Explanation:

i dont know

mart [117]3 years ago
3 0

Answer:

its B

Explanation:

i just took the test

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A 200N force pushes forward on the 20kg truck. The truck pulls two crates connected together by ropes, as shown. The gravitation
devlian [24]

Force on left crate, right crate and truck we need to draw here

First we will write the equation for left crate

T_{left} = m*a

T_{left} = 20*a

Now similarly for right crate

T_{right} - T_{left} = 20*a

now for truck

F - T_{right} = 20*a

Now we know that F = 200 N

200 - T_{right} = 20*a

now add all three equation

T_{left} + T_{right} - T_{left} + 200 - T{right} = 20a + 20 a + 20a

200 = 60 a

a = 10/3 m/s^2

now we will find all tension using this value of acceleration

T_{left} = 20 * \frac{10}{3} = \frac{200}{3} N

T_{right} - \frac{200}{3} = 20*\frac{10}{3}

T_{right} = \frac{400}{3} N


6 0
3 years ago
A ski lift has a one-way length of 1 km and a vertical rise of200m. The chairs are spaced 20 m apart, and each chair can seatthr
Strike441 [17]

Answer:

a) 68.125 KW

b) 43.04 KW

Explanation:

Distance =d= 1 km

Height = h = 200 m

Spacing between chairs =D= 20 m

No. of people per chair = 3

Speed = V= 10 km/h= 10000m\3600 s=2.8 m/s

mass of chair = 250 kg

Work to operate sky lift

W= mgh

Number of chairs any moment operational = N= 1 km/20=1000m/20=50

So, total mass of chairs = 50 × 250 =12500kg

so, w= mgh=12500×9.8×200=2452500 j

Power is rate of work - we need time for operation time of lift

Time= t = distance/speedf= 1km/(10km/h)=1km/(10 km/3600s)=360s

So, Power P= Work/time=w/t=12500/360=68125 j/s=68.125 KW

Now calculating power for operating speed in 5 sec

We calculate accelleration=a for 5 sec

a= speed / time= V/52.8/5=0.28 m/sec2

for vertical acceleration we calculate θ angle first;

tanθ= height /distance= 200/1000= 0.2

==> θ=11.34°

Vertical acceleration = a₁=a sinθ= 0.28× sin 11.34=0.10835 m/sec2

to calculate height gained during startup use;

S=vit+1/2at2

here s=H

vi=0m/s

t=5 sec

==> H = 0+1/2a₁×t=0.5×0.10835 ×5²=0.5×0.10835×5×5=1.362 m

Total Work =mgH+0.5×m×V²=12500(9.8×1.362+0.5×2.8×2.8)=215240.56 j

Again power = work / time=215240.56/5=43048.112 W=43.04 KW

6 0
2 years ago
2 examples of balanced forces
valina [46]

Answer:

<h2>Here are some examples of situations involving balanced forces. </h2><h2>Hanging objects. The forces on this hanging crate are equal in size but act in opposite directions.</h2><h2>Floating in water. Objects float in water when their weight is balanced by the upthrust from the water.</h2><h2>Standing on the ground.</h2>

I hope this helps

6 0
3 years ago
How much potential energy does an 8 kg flower pot hanging 5 m above the ground have
Morgarella [4.7K]

Answer:

Explanation:

Relative to ground level it has

PE = mgh = 8(10)(5) = 400 J

8 0
2 years ago
Read 2 more answers
Where on this diagram does the ball have the highest point of gravitational potential energy?
mixer [17]
It should be at the very top since it has more space to fall which gives it more potential energy
3 0
2 years ago
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