Ask question

Ask question

All categories

katovenus [111]

3 years ago

9

A rock rolls down a steep hill. Its intial velocity is 1 meter per second. By the time it reaches the bottom of the hill 30 seco

nds later. Its velocity is 7 meters per second. What is the acceleration of the rock

1 answer:

Andru [333]3 years ago

7 0

2 meters per second is the answer

You might be interested in

The weight of a luggage is 69.3 N on the moon. Find its weight on the Earth.

Galina-37 [17]

Answer:

Explanation:

weight on moon = 1/6* weight on earth

69.3=1/6*weight on earth

weight on earth = 69.3*6

weight on earth = 415.8 N

8 0

3 years ago

What is the explanation for how a modern transmission electron microscope (TEM) can achieve a resolution of about 0.2 nanometers

IgorC [24]

Answer:

Explanation:

A simple light microscope uses light for imaging of objects where as a transmission electron microscope uses a monochromatic beam of electrons.

This beam is passed by a magnetic field which is very strong and thus act as a lens.

Its resolution of very high which is about 0.2 nanometers because of the separation between two atoms.

Because of this reason its resolution is about 1000 times greater than light microscope.

3 0

3 years ago

How much would a 15.0 kg object weigh on Neptune?

yan [13]

The gravity on Neptune is 11.15 m/s²
the gravity on earth is 9.81 m/s²
divide the Neptune and earth gravity we get 1.13
which means object on neptune is 1.13 heavier than earth
yield, weigh of the object on neptune is 1.13×15=17.04kg

5 0

3 years ago

Read 2 more answers

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

4 years ago

A student makes a model of the sun-Earth system by swinging a ball around her head. Using this model, the student is trying to e

DedPeter [7]

Answer:

c. gravitational attraction between the sun and earth

8 0

3 years ago

Read 2 more answers