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lilavasa [31]
2 years ago
5

A box of mass m is pulled with a constant acceleration a along a horizontal frictionless floor by a wire that makes an angle of

15° above the horizontal. If T is the tension in this wire, then _____________.
A) T = ma.
B) T > ma.
C) T < ma.
Physics
1 answer:
emmainna [20.7K]2 years ago
6 0

Answer:

B) T > ma.

Explanation:

To solve this problem, we have to analyze the forces acting in the horizontal direction.

In the horizontal direction, we have:

- The horizontal component of the tension in the wire, Tcos \theta, where T is the magnitude of the tension and \theta the angle that the wire makes with the horizontal

Since this is the only force acting on the box in the horizontal direction, this is also the net force, so it is equal to the product of mass and acceleration (Newton's second law of motion):

Tcos \theta = ma

where

m is the mass of the box

a is the acceleration

We can rewrite the equation as

T=\frac{ma}{cos \theta}

The angle in this problem is \theta=15^{\circ}, so

T=\frac{ma}{cos 15^{\circ}}=\frac{ma}{0.966}=1.035 ma

Therefore, the correct option is

B) T > ma.

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4 0
2 years ago
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1. A student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.
Snezhnost [94]

Explanation:

(a) Displacement of an object is the shortest path covered by it.

In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.  She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.

0.4 miles = 0.64 km

displacement = 0.7-0.3+0.64 = 1.04 km

(b) Average velocity = total displacement/total time

t = 15 min = 0.25 hour

v=\dfrac{1.04\ km}{0.25\ h}\\\\v=4.16\ km/h

Hence, this is the required solution.

8 0
2 years ago
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

Charge on proton,q=1.6\times 10^{-19} C

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
2 years ago
A typical adult human has a mass of about 70 kg.
Misha Larkins [42]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2 
<span>= 2.40 x 10-3 N

b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N 
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3 0
2 years ago
Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per
vfiekz [6]

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum  theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work  and amount of entropy  and magnitude

solution

first we calculate work i.e heat transfer - work =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C    .................1

so first we get some value from steam table with the help of 100 bar @360°C and  1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change =  specific entropy ( 100 bar @360°C)  - specific entropy ( 1 bar @160°C )

put these value we get

entropy change =  7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum  theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum  theoretical work = specific internal energy - 2038.3

maximum  theoretical work = 2728 - 2038.3

maximum  theoretical work is 689.4 kJ/kg

3 0
2 years ago
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