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Elden [556K]
2 years ago
8

Block 1 and Block 2 have the same mass, m, and are released from the top of two inclined planes of the same height making 30 deg

ree and 60 degree angles with the horizontal direction, respectively. If the coefficient of friction is the same in both cases, which of the blocks is going faster when it reaches the bottom of its respective incline?
a. Block 1 is faster.
b. Block 2 is faster.
c. We must know the actual masses of the blocks to answer.
d. Both blocks have the same speed at the bottom.
e. There is not enough information to answer the question because we do not know the value of the coefficient of kinetic friction.
Physics
1 answer:
steposvetlana [31]2 years ago
5 0
I believe c is the answer and if not it is e
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Stella [2.4K]
Hello <span>Andijwiltbank 
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Question: <span>Often what one expects to see influences what is perceived in the surrounding environment. True or False?

Answer: True

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8 0
2 years ago
Read 2 more answers
Is v2 = v1t+a dimensionally correct? Explain please!
Lady bird [3.3K]
You want v2 = v1 + at
v is measured in m/s, a in m/s2, and t in s.
the dimensions multiply like algebraic quantities. 
so because v2 is measured in m/s, then (v1 + at) has to come out in m/s
 the units for (v1 + at) are (m/s) + (m/s2)(s)
time "s" cancels out one acceleration "s", so it comes ut to (m/s) + (m/s), which = (m/s). 
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4 0
2 years ago
A 250 Hz tuning fork is struck and the intensity at the source is I1 at a distance of one meter from the source. (a) What is the
Zina [86]

Answer:

a) 0.0625 I_1

b) 3.16 m

Explanation:

<u>Concepts and Principles  </u>

The intensity at a distance r from a point source that emits waves of power P is given as:  

I=P/4π*r^2                         (1)

<u>Given Data</u>

f (frequency of the tuning fork) = 250 Hz

I_1 is the intensity at the source a distance r_1 = I m from the source.  

<u>Required Data</u>

- In part (a), we are asked to determine the intensity I_2 a distance r_2 = 4 in from the source.

- In part (b), we are asked to determine the distance from the tuning fork at which the intensity is a tenth of the intensity at the source.  

<u>solution:</u>

(a)  

According to Equation (1), the intensity a distance r is inversely proportional to the distance from the source squared:

I∝1/r^2

Set the proportionality:  

I_1/I_2=(r_2/r_1)^2                                 (2)

Solve for I_2 :  

I_2=I_1(r_2/r_1)^2  

I_2=0.0625 I_1

(b)  

Solve Equation (2) for r_2:  

r_2=(√I_1/I_2)*r_1

where I_2 = (1/10)*I_1:

r_2=(√I_1/1/10*I_1)*r_1

     =3.16 m

3 0
2 years ago
A 1500 W radiant heater is constructed to operate at 115 V. (a) What will be the current in the heater? (b) What is the resistan
OlgaM077 [116]

Answer:

a) I = 13.04 A

b)  R = 8.82 ohms

c) 1291.87 kilocalories are generated an hour.

Explanation:

let P be the power of the heater, V be the voltage of the heater, I be the current of the heater, R be the resistance.

a) we know that:

P = I×V

I = P/V

  = (1500)/(115)

  = 13.04 A

Therefore, the current of the heater is 13.04 A

b) we now have voltage and current, according to Ohm's law:

R = V/I

  = (115)/(13.04)

  = 8.82 ohms

Therefore, the resistance of the heating coil is 8.82 ohms.

c) the number of kilocalories generated in one hour by the heater is just the energy the heater produces in one hour which is given by:

E = P×t

  = (1500)(1×60×60)

  = 5400000 J

since 1 calorie = 4.81 J

1 kilocalorie = 0.001 calories

E = 5400000/4.18 ≈ 1291866.029 calories ≈1291.87 kilocalories

Therefore, 1291.87 kilocalories are produced/generated in one hour.

8 0
2 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
2 years ago
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