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Anton [14]
4 years ago
10

Please solve this problem inelastic collision

Physics
1 answer:
S_A_V [24]4 years ago
7 0

It will crash at a speed of 5 m/s and that's basically it.

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A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with
NNADVOKAT [17]

Answer: 1.59 sec

Explanation:

The kinetic energy contained in the rotation of the cylinder is 1/2 m v^2

The kinetic energy of translation is also 1/2 m v^2

so the total energy is m v^2

The force applied is mg sin (theta)

= m x 9.8 x 1/2

= 4.9 m

Now equate

F x d = m v^2

4.9 m x 3.1 = m v^2

v^2 = 4.9 x 3.1

v = sqrt(4.9 x 3.1) = 3.9 m/s

Acceleration.

V^2 = 2 a s

4.9 x 3.1 = 2 x a x 3.1

4.9 = 2 a

a = 2.45 m/s^2

Time

T = v/a

= 3.9/2.45

= 1.59 sec

6 0
4 years ago
Which of these is another name for Newton's
dezoksy [38]
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.
5 0
3 years ago
Read 2 more answers
The length of the mercury thread is found to be 4cm and 24cm at ice point and steam point respectively on an ungraduated thermom
BabaBlast [244]

Answer:

The difference between ice and steam in Celsius (Centigrade) is 100 deg.

So the difference between and 4 cm and 24 cm of the thread corresponds to 100 deg C.

So 8 cm is 4 cm greater than the ice point

4 cm / 20 cm = 1/5     since the steam point and the ice point are 20 cm apart

Then 1/5 * 100  deg C = 20 deg C   the requested temperature

6 0
3 years ago
A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continue
enot [183]

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

7 0
3 years ago
Which of the following represents an upright image? <br> A. -do<br> B. +m<br> C. -m<br> D. +do
Tom [10]

Answer:

B. +m

Explanation:

The magnification of an image is defined as the ratio between the size of the image and of the object:

m = \frac{y'}{y}

where we have

y' = size of the image

y = size of the object

There are two possible situations:

- When m is positive, y' has same sign as y: this means that the image image is upright

- When m is negative, y' has opposite sign to y: this means that the image is upside down

Therefore, the correct option representing an upright image is

B. +m

5 0
3 years ago
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