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lilavasa [31]
3 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]3 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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3 years ago
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1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g
Delvig [45]

1) 6.67\cdot 10^{-4} N

The force of gravitation between the two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2} is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N


2)  2.7\cdot 10^{-3} N

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

F \sim \frac{1}{r^2}

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F

So, the gravitational force increases by a factor 4. Therefore, the new force will be

F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N


3)  12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

\tau=Fd

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

\tau=(25 N)(0.5 m)=12.5 Nm


4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity (\omega) is:

L=I\omega

In this problem, we have

I=0.007875 kgm^2

\omega=6.28 rad/s

So, the angular momentum is

L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s

6 0
3 years ago
An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?
hjlf

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

power P= 100W

Required

The current I

Step two

Yet this power is also given by

P = I^2R

make I subject of the formula we have

I= \sqrt{\frac{P}{R} }

substitute

I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps

8 0
3 years ago
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