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Kitty [74]
3 years ago
14

I need help please and thank you

Physics
1 answer:
EastWind [94]3 years ago
7 0

Answer: magnitude= 2 m/s/s

Explanation: I goes up 2 every second

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In the case of a swinging pendulum, Potential Energy is greatest when ____.
GREYUIT [131]
I am almost 100% positive the answer is A.
4 0
3 years ago
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A pipe branches symmetrically into two legs of length L, and the whole system rotates with angular speed ω around its axis of sy
andre [41]

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

7 0
3 years ago
The wall of a large room is covered with acoustic tile in which small holes are drilled 4.6 mm from center to center. How far ca
Zielflug [23.3K]

Answer:

L= 27.42m

Explanation:

We have the variables

D=4.6*10^{-3}m

d= 4*10^{-3}m

\lambda = 550*10^{-9}m

Use the relation,

L=\frac{D}{\theta_R} = \frac{D_d}{1.22\lambda}

L= \frac{(4.6*10^{-3}m)(4*10^{-3}m)}{1.22(550*10^{-9}m)}

L= 27.42m

8 0
4 years ago
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle res
zhannawk [14.2K]

Answer:

a) 250 N/m

b) 22.4 rad/s , 3.6 Hz , 0.28 sec

c) 0.3125 J

Explanation:

a)

F = force applied on the spring = 7.50 N

x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m

k = spring constant of the spring

Since the force applied causes the spring to stretch

F = k x

7.50 = k (0.03)

k = 250 N/m

b)

m = mass of the particle attached to the spring = 0.500 kg

Angular frequency of motion is given as

w = \sqrt{\frac{k}{m}}

w = \sqrt{\frac{250}{0.5}}

w = 22.4 rad/s

f = frequency

Angular frequency is also given as

w = 2 π f

22.4 = 2 (3.14) f

f  = 3.6 Hz

T = Time period

Time period is given as

T = \frac{1}{f}

T = \frac{1}{3.6}

T = 0.28 sec

c)

A = amplitude of motion = 5 cm = 0.05 m

Total energy of the spring-block system is given as

U = (0.5) k A²

U = (0.5) (250) (0.05)²

U = 0.3125 J

5 0
4 years ago
A 49.5 gram super ball traveling at 25.5 m/s bounces off a brick wall and rebounds at 19.5 m/s. a high-speed camera records this
OLga [1]
T = 4.25 ms = 4 x 10⁻³ s, the time for rebound
v₁ = 25.5 m/s, the impacting velocty
v₂ = -19.5 m/s, the rebounding velocity (n the opposite directon)

The change in velocity is
v₂ - v₁ = - (25.5+19.5) = -45 m/s

The acceleration is
a = (-45 m/s)/(4 s) = -11.25 m/s²
The negative sign indicates that the final velocity is opposiye to the impact velocty.

Answer: The magnitude of the acceleration is 11.25 m/s²
8 0
3 years ago
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