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3 years ago

How does using a ramp to load boxes into a truck make work easier?

2 answers:

8 0

"Using the ramp decreases the amount of force needed to move the boxes, but the boxes must be moved over a longer distance" is the way among the following choices given in the question that a ramp <span>to load boxes into a truck make work easier. The correct option among all the options that are given in the question is option "D".</span>

larisa [96]3 years ago

3 0

The correct answer is

D. Using the ramp decreases the amount of force needed to move the boxes, but the boxes must be moved over a longer distance.

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A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed

Liono4ka [1.6K]

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

$F = \dfrac{mv^2}{r}$

m is the mass, v is the velocity and r is the radius.

It follows that $F \propto v^2$ , provided m and r are constant.

When v is doubled, the new force, $F_1$ , is

$F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F$

Hence, the tension in the string is quadrupled.

8 0

3 years ago

How many nanoseconds are there in 1.90 yr ?<br> Express your answer using three significant figures.

shepuryov [24]

   (1.9 yr) x (365.24 day/yr) x (86,400 sec/day) x (10⁹ nsec/sec)

   = (1.9 x 365.24 x 86,400 x 10⁹) nanosec

   = 6.00 x 10¹⁶ nanoseconds

5 0

3 years ago

Are you country or are you city?

Mariulka [41]

Answer:

Neither lma0 I'm from a town :P

Explanation:

Hbu?

Have a nice dayyy <3

6 0

3 years ago

Read 2 more answers

Your shopping cart has a mass of 65 kilograms, In order to accelerate the shopping cart down an aisle at 0.3 m/sec2, what force

garri49 [273]

Remember Newton's second law: F=ma

to get the force in newtons, mass should be in kg and acceleration in m/s^2

conveniently, we don't need to convert units

we just need to multiply the two to get the force

65* 0.3 = 19.5 kg m/s^2 or N

if significant digit is an issue, the least number if sig figs is 1 so the answer would be 20 N

4 0

3 years ago

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

2 years ago