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blondinia [14]
3 years ago
9

In straight line motion, if the velocity of an object is changing at a constant rate, then its position is _________ and its acc

eleration is___________: O changing: zero O changing; changing O constant and non-zero; constant and non-zero O None of the above
Physics
1 answer:
Serjik [45]3 years ago
8 0

Answer:

None of the above

It should be position is changing and acceleration is constant.

Explanation:

Since the velocity is changing, this means the object is moving, so the <u>position must also be changing</u>.

Acceleration is the change in velocity in time, if this change of velocity happens at a constant rate, the <u>acceleration must be constant</u> too.

So, for example, if the velocity were to stay the same (not changing), acceleration would be zero, because there wouldn't be a change in time on the velocity.

So in this case the answer sould be position is changing and acceleration is constant. But this isn't in the options so the correct answer is "None of the above"

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If you walk at an average speed of 5 km/h for 30 minutes, how
Inessa05 [86]

The distance that would be accumulated during the journey is 2.5 meters

The parameters given in the question are  written below;

average speed= 5 km/hr

time = 30 minutes

convert 30 minutes to hours

= 30/60

= 0.5 hours

Distance-= speed × time

= 5 × 0.5

= 2.5 meters

Hence the distance of the entire journey is 2.5 meters

Please see the link below for more information

brainly.com/question/24268730?referrer=searchResults

3 0
2 years ago
At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

V - Sample volume, measured in cubic meters.

P - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0
3 years ago
Sitting in a chairlift, Rebecca has a gravitational potential energy of 5,997.6
stira [4]

Answer:

B) 12 m

Explanation:

Gravitational potential energy is:

PE = mgh

Given PE = 5997.6 J, and m = 51 kg:

5997.6 J = (51 kg) (9.8 m/s²) h

h = 12 m

8 0
3 years ago
The fluid pressure 10 ft underwater is _____ the fluid pressure 5 ft underwater. A.less than or greater than, depending on what
just olya [345]
<span>I would say greater than because as you do deeper, the pressure strengthens. If you were in a 10 ft deep pool and you dive all the way to the bottom, the ears usually pop. That's because of the pressure. Whereas if you were to go five feet, your ears wouldn't. It depends on the age of the person. Hope this helps.</span>
5 0
3 years ago
A far-sighted person has a near-point of 80 cm. To correct their vision so that they can see objects that are as close as 10 cm
bekas [8.4K]

Answer:

f = 8.89 cm

Explanation:

As we know that Far sighted person has near point shifted to 80 cm distance

so he is able to see the object 80 cm

now the distance of lens from eye is 2 cm

and the person want to see the objects at distance 10 cm

so here the image distance from lens is 80 cm and the object distance from lens is 8 cm

now from lens formula we have

\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}

-\frac{1}{80} + \frac{1}{8} = \frac{1}{f}

f = 8.89 cm

3 0
3 years ago
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