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kramer
3 years ago
13

A 60 kg student is standing atop a spring in an elevator that is accelerating upward at 3.0 m/s2. The spring constant is 2.5 x 1

03 N/m. By how much is the spring compressed?
Physics
1 answer:
dimaraw [331]3 years ago
6 0

the spring will be compressed by 0.3072 m

Explanation:

acceleration of elevator=3 m/s²

mass of student= 60 Kg

spring constant=2.5 x 10³ N/m

the force on the student is given by F = m ( g +a)

F=60 (9.8+3)

F=768 N

now the formula for spring force is given by

F= k x

768= 2.5 x 10³ (x)

x=0.3072 m

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Order the steps to describe how information is sent. A signal is produced. A signal is seen, heard, or used. Radio waves are mod
nordsb [41]

Answer:

A signal is produced.

Radio waves are modulated and amplified.

Radio waves are amplified and demodulated.

A signal is seen, heard, or used.

Explanation:

Communication is the method of transmitting information between two or more participants. A radio communication network is a set of static and wireless radio equipment developed to facilitate an entity by enabling particular modes of communication such as one-to-many and one-to-one communication. Radio waves are being used to transmit information spatially from the sender to receiver, by modulating the radio signal in the transmitter. The above steps explains the loop of the radio communication system.

6 0
3 years ago
Which subatomic particle has a positive charge?
VMariaS [17]
The answer is protons
Electrons have negative charge and neutrons have 0 charge
6 0
3 years ago
Read 2 more answers
A 6 kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction μk=0.15 by a co
ozzi

Answer:

1.8 m/s

Explanation:

Draw a free body diagram of the block.  There are four forces:

Normal force Fn up.

Weight force mg down.

Applied force F to the east.

Friction force Fn μ to the west.

Sum the forces in the y direction:

∑F = ma

Fn − mg = 0

Fn = mg

Sum the forces in the x direction:

F − Fn μ = ma

F − mg μ = ma

a = (F − mg μ) / m

a = (12 N − 6 kg × 9.8 m/s² × 0.15) / 6 kg

a = 0.53 m/s²

Given:

Δx = 3 m

v₀ = 0 m/s

a = 0.53 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (0.53 m/s²) (3 m)

v = 1.8 m/s

8 0
3 years ago
A ball is thrown directly downward with an initial speed of 7.65 m/s froma height of 29.0 m. After what time interval does it st
coldgirl [10]

Answer: 1.77 s

Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:

xf= xo+vo*t+(g*t^2)/2  we can consider the origin on the top so the xo=0 and xf=29 m; then

(g*t^2)/2+vo*t-xf=0  vo is the initail velocity, vo=7.65 m/s

then by solving the quadratric equation in t

t=1.77 s

8 0
3 years ago
Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges
Nostrana [21]

Answer:

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

E_1 = E_2

Let the position where net field is zero will lie at distance "r" from q1

\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}

now we will have

\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}

now square root both sides

\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}

now we have

\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1

so we have

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

8 0
3 years ago
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