Ask question

Ask question

All categories

3 years ago

15

You're conducting a physics experiment on another planet. You drop a rock from a height of 2.1 m and it hits the ground 0.6 seco

nds later. What is the acceleration due to gravity on this planet?
A. 7.0 m/s2
B. 9.8 m/s2
C. 11.7 m/s2
D. 5.2 m/s2

(What formula should I use to solve this on my own?)

2 answers:

WINSTONCH [101]3 years ago

8 0

Answer:

11.7 m/s2

Explanation:

Apex ; good luck everyone! :)

umka21 [38]3 years ago

3 0

The formula to use is the one that connects the acceleration,
the distance fallen, and the time spent falling:

   Distance = 1/2 a T² .

You said 2.1 meters in 0.6 second .

   2.1 m       = 1/2 a (0.6 sec)²

Multiply each side by 2 :       4.2 m =    a (0.6 sec)²

Divide each side by (0.6 sec)² =    (4.2/0.36) m/s² = a

   a = (11 and 2/3) m/s²

(about 19% more than Earth's gravity)

You might be interested in

an object weighing 15 newtons is lifted from the ground to a height of 0.22 meter what is the increase in the object's gravitati

kicyunya [14]

GPE= weight•height= 15 N• 0.22meter= 3.3 Joules
I hope this helps ~~Charlotte~~

5 0

2 years ago

Read 2 more answers

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

2 years ago

A metal pot is filled with water and placed on top of a stove. As the stove heats up, the water begins to boil, and steam can be

stepladder [879]

This type of system is an example of open system because the vapors leave the system and goes into the atmosphere.

<h3>What is open system?</h3>

An open system is a type of system that has external interactions which means energy, or material transfers into or out of the system.

So we can conclude that this type of system is an example of open system because the vapors leave the system and goes into the atmosphere.

Learn more about system here: brainly.com/question/14323743

#SPJ1

4 0

1 year ago

Help please I'll mark as brainliest for the first correct answer

Alenkinab [10]

Answer:

first one i think is this. work = 1/2 kx^2 = 1/2 Fx

2nd, is 0.08 J

Explanation:

EE = ½ kx²

EE = ½ (400 N/m) (0.02 m) ²

EE = 0.08.

THIRD, Velocity of the stone is 4 m/s when it leaves catapult.

3 0

3 years ago

You perform 556 J of work lifting a box to a height of 1.3m. How much force did you use to lift the box?

babunello [35]

The work is equal to the product between the force applied and the distance covered by the box:
$W=Fd$
In our problem, W=556 J, and d=1.3 m (the box is lifted to a height of 1.3 m, so it covered 1.3 m from its initial point). Therefore we can find the force applied to lift the box:
$F= \frac{W}{d}= \frac{556 J}{1.3 m}=427.7 N$

4 0

3 years ago

Read 2 more answers