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2 years ago

Hey besties, im having a mental breakdown on this question. pls help :D

1 answer:

8 0

With "toward the wall" designated as the positive direction, the average acceleration felt the ball in the 0.0158 s of contact with the wall is

$a_{\rm ave} = \dfrac{-26.412\frac{\rm m}{\rm s} - 28.4\frac{\rm m}{\rm s}}{0.0158\,\mathrm s} ≈ \boxed{-3470 \dfrac{\rm m}{\mathrm s^2}}$

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Alexis is a scientist who is studying solid-state physics. Which activity would she most likely do as a part of this research? O

Ann [662]

Answer:

pretty sure its studying the atomic structure of a solid carbon dioxide. so c

Explanation:

8 0

2 years ago

An woman whose weight is 804 N stands on a long horizontal plank of wood 1.55 m from one end. The plank is uniform and is suppor

meriva

Answer:

Explanation:

There will be reaction force by each vertical post on horizontal plank . Let it be R₁ and R₂ . R₁ is reaction force by the post nearer to woman

Taking torque of all forces about the end far away from the woman

Torque by reaction force = R₁ x 5.5

= 5.5 R₁ upwards

Torque by weight of woman in opposite direction , downwards

= - 804 x ( 5.5 - 1.55 )

= - 3175.8

Torque by weight of the plank in opposite direction , downwards .

= - 27 x 5.5 / 2

= - 74.25

Torque by R₂ will be zero as it passes through the point about which torque is being taken .

Total torque

= 5.5 R₁ - - 3175.8 - - 74.25 = 0 ( For equilibrium )

5.5 R₁ = 3250

R₁ = 590.9 N .

6 0

3 years ago

A hockey puck moving at 0.4600 m/s collides with another puck that was at rest. The pucks have equal mass. The first puck is def

Sladkaya [172]

Answer:

Speed =0.283m/ s

Direction = 47.86°

Explanation:

Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane

MU1 =MU2cos38 + MV2cos y ...x plane

0 = MU2sin38 - MV2sin y .....y plane

Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2

Substitute into equation above

.46 = .34cos38 + V2cos y ...equ1

.34sin38 = V2sin y...equ2

.19=V2cos Y...x

.21=V2sin Y ...y

From x

V2 =0.19/cost

Sub V2 into y

0.21 = 0.19(Sin y/cos y)

1.1052 = tan y

y = 47.86°

Sub Y in to x plane equ

.19 = V2 cos 47.86°

V2=0.283m/s

7 0

3 years ago

What is the current in a series circuit that has two resistors (4.0 ohms and

Maslowich

C 1.6 amps hipe this helps

6 0

3 years ago

(b) If you decrease the length of the pendulum by 25%, how does the new period TN compare to the old period T?

malfutka [58]

Answer:

The new period will be reduced by 50%

Explanation:

The period of pendulum is given by;

$T= 2\pi\sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g} }\\\\(\frac{T}{2\pi} )^2 = {\frac{L}{g}}\\\\\frac{T^2}{4\pi ^2} = {\frac{L}{g}}\\\\T^2(\frac{g}{4\pi ^2}) = L\\\\ \frac{g}{4\pi ^2}= \frac{L}{T^2}\\\\\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}$

When the length is decreased by 25%, the new length L₂ is given by;

L₂ = 25/100(L₁)

L₂ = 0.25L₁

$\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{T_1^2L_2}{L_1} \\\\T_N^2 = \frac{T^2(0.25L_1)}{L_1}\\\\ T_N^2 =0.25T^2\\\\T_N = \sqrt{0.25T^2}}\\\\T_N = 0.5 T$

Thus, the new period will be reduced by 50%

8 0

3 years ago