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Rufina [12.5K]
2 years ago
5

Consider the free-body diagram. If you want the box to move, the force applied while dragging must be greater than the

Physics
1 answer:
VLD [36.1K]2 years ago
6 0

You would want it to be greater than D. friction force

It needs be greater than the friction applied to it.

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Do atoms ever touch​
Mumz [18]

Answer:

nope never

Explanation:

8 0
3 years ago
Read 2 more answers
A security guard walks at a steady pace traveling 170 m in one trip around the perimeter of a building.
WARRIOR [948]
Speed= distance/time
Speed= 170/230
Speed=0.74 m/s
6 0
3 years ago
A juggler throws a ball straight up into the air. The ball remains in the air for a time (t) before it lands back in the juggler
natka813 [3]

Answer:

9.8 m/s^2, downward

Explanation:

There is only one force acting on the ball during its motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity (downward)

According to Newton's second law,

F=ma

where F is the net force on the object and a is its acceleration. Rearranging for a,

a=\frac{F}{m}

As we said, the only force acting on the ball is gravity, so F = mg and the acceleration of the ball is:

a=\frac{mg}{m}=g

Therefore, the ball has a constant acceleration of 9.8 m/s^2 downward for the entire motion.

5 0
3 years ago
Which will have a larger momentum when moving at the same speed: a 2,000-kg truck or a 1,000-kg sedan
Alina [70]

Answer:

2000 kg

Explanation:

Given that Which will have a larger momentum when moving at the same speed: a 2,000-kg truck or a 1,000-kg sedan

According to the definition of momentum, momentum is the product of mass and velocity.

That is,

Momentum = mass × velocity

Since velocity or speed is the same, then, the one of higher mass will have a greater momentum.

Therefore, the 2000 kg truck will have the greater momentum.

5 0
2 years ago
If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
kotegsom [21]

Answer:

0.82 mm

Explanation:

The formula for calculation an n^{th} bright fringe from the central maxima is given as:

y_n=\frac{n \lambda D}{d}

so for the distance of the second-order fringe when wavelength \lambda_1 = 745-nm can be calculated as:

y_2 = \frac{n \lambda_1 D}{d}

where;

n = 2

\lambda_1 = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y_2 = 0.00276 m

y_2 = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y^'}_2 = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

\delta y =  0.82 × 10 ⁻³ m (\frac{1.0mm}{10^{-3}m} )

\delta y = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

6 0
3 years ago
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