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Sergio039 [100]
3 years ago
15

7) T F If two forces of equal magnitude act on an object that is hinged at a pivot, the force acting farther from the pivot must

produce the greater torque about the pivot.
Physics
1 answer:
saul85 [17]3 years ago
4 0

Answer:

False

Explanation:

The torque exerted by a force is given by:

\tau=Fd sin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the pivot

\theta is the angle between the directions of F and d

We see that the magnitude of the torque depends on 3 factors. In this problem, we have 2 forces of equal magnitude (so, equal F). Moreover, one of the forces (let's call it force 1) acts farther from the pivot than force 2, so we have

d_1 > d_2

However, this does not mean that force 1 produces a greater torque. In fact, it also depends on the angle at which the force is applied. For instance, if the first force is applied parallel to d, then we have

\theta_1 =0\\sin \theta=0

and the torque produced by this force would be zero.

So, the statement is false.

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A car starts from rest and after 20 seconds it's velocity becomes 108km find the acceleration of the car
andre [41]

Answer:

1.5 km/s²

Explanation:

Given that:

a car starts from rest; it means the initial velocity (u) = 0 km/hr = 0 m/s

after time (t) = 20 seconds

the final velocity = 108 km/hr = 30 m/s

The acceleration (a) of the car can be determined by using the formula:

a = \dfrac{v-u}{t}

a = \dfrac{30\  m/s -0 \ m/s}{20 \ s}

a = \dfrac{30 \  m/s}{20 \ s}

a = 1.5 km/s²

7 0
3 years ago
Can you respond this two questions, please? :
Andrews [41]

Where's the diagram for question 1?

6 0
3 years ago
A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me tempe
Simora [160]

Answer:

Final temperature is 295K

Explanation:

Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C×\frac{1mol}{63,546g}×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C×\frac{1mol}{18,02g}×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK×\frac{1mol}{63,546g}×25,0g×(X-363K) = -75,2J/molK×\frac{1mol}{18,02g}×100,0g× (X-293K)

9,64X J/K - 3499J = - 417X J/K + 122273J

426,64X J/K = 125772 J

<em>X = 295K</em>

<em></em>

Final temperature is 295K

I hope it helps!

6 0
3 years ago
A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its
gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

d = 2\pi r ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
v = \sqrt{\frac{Gm}{r}}

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

7 0
2 years ago
A truck traveling at a constant speed of 28 m/s passes a more slowly moving car. The instant the truck passes the car, the car b
choli [55]

Answer:

the velocity of car when it passes the truck is u = 16.33 m/s

Explanation:

given,

constant speed of truck  = 28 m/s

acceleration of car = 1.2 m/s²

passes the truck in 545 m

speed of the car when it just pass the truck = ?

time taken by the truck to travel 545 m

              time =\dfrac{distance}{speed}

              time =\dfrac{545}{28}

              time =19.46 s

velocity of the car when it crosses the truck

S = ut + \dfrac{1}{2}at^2

545= u\times 19.46 + \dfrac{1}{2} \times 1.2 \times 19.46^2

u = 16.33 m/s

the velocity of car when it passes the truck is u = 16.33 m/s

5 0
3 years ago
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