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3 years ago

7Which of the following terms describes how glaciers move?

Physics

2 answers:

tatuchka [14]3 years ago

7 0

Answer:

D is the answer I think (0 w 0 )

Explanation:

ollegr [7]3 years ago

5 0

Your answer should me D

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To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circus performer wh

sp2606 [1]

Answer:

$v = 15.45 m/s$

Explanation:

As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have

Work done by friction force = Loss in mechanical energy

so we will have

$W_f = (U_i + K_i) - (U_f + K_f)$

here we know that

$W_f = F_f . d$

$W_f = 40 \times 4$

$W_f = 160 J$

Initial compression in the spring is given as

$F = kx$

$4400 = 1100 x$

$x = 4 m$

now from above equation

$W_f = (\frac{1}{2}kx^2 + 0) - (mgh + \frac{1}{2}mv^2)$

$160 = (\frac{1}{2}1100(4^2) + 0) - (60 \times 9.8\times 2.50 + \frac{1}{2}(60)v^2)$

$160 = 8800 - 1470 - 30 v^2$

$v = 15.45 m/s$

3 0

3 years ago

The standard measure used to compare sound intensities is the ______ .

uranmaximum [27]

The answer is Decibels. <span />

4 0

3 years ago

Read 2 more answers

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

il63 [147K]

Answer:

0.0000109261200583 s

0.0109261200583

Explanation:

$d_2$ = Distance from right ear = 3 m

s = Distance between ears = 15 cm

v = Speed of sound in air = 343 m/s

Distance between the left ear and the bird

$d_1=\sqrt{s^2+d_2^2}\\\Rightarrow d_1=\sqrt{0.15^2+3^2}\\\Rightarrow d_1=3.00374765918\ m=3.004\ m$

Time

$t=\dfrac{Distance}{Speed}$

Time difference would be

$\Delta T=\dfrac{d_1}{v}-\dfrac{d_2}{v}\\\Rightarrow \Delta T=\dfrac{3.00374765918}{343}-\dfrac{3}{343}\\\Rightarrow \Delta T=0.0000109261200583\ s$

The time difference is 0.0000109261200583 s

Time period is given by

$T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1000}\\\Rightarrow T=10^{-3}\ s$

The ratio is

$\dfrac{\Delta T}{T}=\dfrac{0.0000109261200583}{10^{-3}}\\\Rightarrow \dfrac{\Delta T}{T}=0.0109261200583$

The ratio is 0.0109261200583

7 0

3 years ago

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0

3 years ago

Suppose snow is falling in Utah. What type of air mass is responsible for this weather?

Viefleur [7K]

Maritime polar air mass comes from the northern part of the pasific ocean and that air mass goes to Utah.

5 0

3 years ago