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2 years ago

What task requires the most work, lifting a 12-kg sack 2 meters or lifting a 25-kg sack 1 meter?

Physics

1 answer:

MrMuchimi2 years ago

5 0

Multiply the masses by the respective distances:

(12 kg) (2 m) = 24 J

(25 kg) (1 m) = 25 J

so the heavier bag takes more work to lift, and (b) is the answer.

(d) is technically correct if the sacks are carrying different contents whose masses are not equal, but since we don't know what's inside each sack, assume 12 kg and 25 kg are the masses of each sack *and* their contents.

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What is the magnitude of g at a height above Earth's surface where free-fall acceleration equals 6.5m/s^2?

prohojiy [21]

You've given the answer, right there in your question.

The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.

We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
the distance from the Earth's center at that height is about

(1 / √0.66) = 1.23 times

the Earth's radius, so the height is about 910 miles above the surface.

7 0

3 years ago

Determine e when I = 0.50 A and R = 12 W.

sammy [17]

Answer:

The correct answer is "24 V".

Explanation:

The given values are:

Current,

I = 0.50 A

Resistance,

R = 12 W

As we know,

⇒ $I = 0.5\times (\frac{E}{2R})$

On substituting the given values, we get

⇒ $0.5= (\frac{E}{4\times 12} )$

⇒ $0.5= (\frac{E}{48} )$

⇒ $E=24 \ V$

7 0

2 years ago

Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is r

Naddik [55]

Explanation:

Fgravity = G*(mass1*mass2)/D²

G is the gravitational constant throughout the universe.

D is the distance between both objects.

D is now reduced by a factor of 5, meaning Dnew = D/5 we get

Fgravitynew = G*(mass1*mass2)/(D/5)² =

= G*(mass1*mass2)/(D²/25) =

= 25* G*(mass1*mass2)/D² = 25* Fgravity

the new force of gravity/attraction is 25×16 = 400 units.

4 0

2 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!<br> I NEED THE ANSWER NOW

FrozenT [24]

90 F = 43 OR 0.9F = 0.43
(F = 43 / 90 OR 0.43 / 0.9 =) 0.48 N

upwards force = downwards force
(R =) 1.2 N

5 0

2 years ago

A 13561 N car traveling at 51.1 km/h rounds

Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

$a_{c} = \frac{v^{2}}{r}$

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

$a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}$

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

$F_{c} = ma_{c}$

Where:

m: is the mass of the car

The mass is given by:

$P = m*g$

Where P is the weight of the car = 13561 N

$m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg$

Now, the centripetal force is:

$F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N$

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

$F_{c} = F_{\mu}$

$F_{c} = \mu N = \mu P$

$\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069$

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!

3 0

3 years ago