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LenaWriter [7]
2 years ago
6

You swing a bat and hit a heavy box with a force of 1500 N. The force the box exerts on the bat is

Physics
1 answer:
vlabodo [156]2 years ago
4 0

Answer:

E)

Explanation:

Because of Newton's Second Law, which says any action will be opposed by a reaction of the same magnitude, then no matter what, no matter if the box does or does not move, the bat bouces back, or what happens after the hit, when hitting the box the force ithbox exerts on the bat is  exactly 1500 N.

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You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac
zaharov [31]

Answer:

Explanation:

Given

Original Frequency f=723\ Hz

apparent Frequency f'=697\ Hz

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

f'=f\cdot \frac{v-v_o}{v+v_s}

where  

f'=apparent frequency  

v=velocity of sound in the given media

v_s=velocity of source

v_0=velocity of observer  

here v_0=0

697=723\cdot (\frac{343-0}{343+v_s})

v_s=(\frac{f}{f'}-1)v

v_s=(\frac{723}{697}-1)\cdot 343

v_s=12.79\approx 12.8\ m/s

i.e.fork acquired a velocity of 12.8 m/s

distance traveled by fork is given by

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v_s^2-0=2\times 9.8\times s

s=\frac{12.8^2}{2\times 9.8}

s=8.35\ m

                                       

5 0
2 years ago
Why a steel ship becomes magnetized as it is constructed?
Inessa [10]
<em>Iron, and to a lesser degree, steel, can only become magnetised by passing an electrical current through it (an electromagnet). So a steel ship does not become magnetised in the accepted sense during construction. </em>

<span><em>However, any large mass of iron will affect the accuracy of a magnetic compass, causing it to deviate wildly from magnetic North. This problem was encountered when iron ships were first constructed in the mid-19 Century. It was overcome by mounting the compass in a 'binnacle', a housing containing two large soft iron balls either side of the compass itself, which counteracted the effect of the hull and balanced the compass so that it read correctly</em></span>
6 0
2 years ago
A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal
zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0
2 years ago
The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -41 oC and 900 Pa, respect
Fittoniya [83]

Answer:

Part a)

\rho = 0.0205 kg/m^3

Part b)

\rho = 1.22 kg/m^3

So density of atmosphere at Martian Surface is very less than the density at Earth.

Explanation:

Part a)

As per ideal gas equation we know that

PM = \rho RT

here we know that Martian atmosphere is equivalent to that of carbon

so we will have

P = 900 Pa

T = 273 - 41 = 232 K

now we will have

(900)(0.044) = \rho (8.31)(232)

\rho = 0.0205 kg/m^3

Part b)

Now for the earth surface the density of air is given for

P = 101.6 kPa

T = 18 ^oC

so we will have

PM = \rho RT

(101.6\times 10^3)(0.029) = \rho(8.31)(273 + 18)

\rho = 1.22 kg/m^3

So density of atmosphere at Martian Surface is very less than the density at Earth.

8 0
2 years ago
You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved
oksian1 [2.3K]

Answer:

Explanation:

The volume of contaminated water

= cross sectional area x height of water level

3.14 x 9 x 9 x 7.5 ft³

= 1907.55 ft³

mass = density x volume

= 1907.55 x 63.5 lbs

m = 121129.425 lbs

This mass has to be raised to the height of 8 ft before evacuation .

There is a rise of centre of mass of

8 - 7.5/2 ft

h = 4.25 ft

Energy required

= mgh

= 121129.425 x 32 x 4.25

= 16473601.8 unit.

3 0
2 years ago
Read 2 more answers
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