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forsale [732]
3 years ago
8

174cm=____ m1.7417.4174017,400​

Physics
1 answer:
FinnZ [79.3K]3 years ago
5 0
Correct answer is 1.74cm
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A 5 kg block is being pushed horizontally across a level surface at a constant velocity. What is the magnitude of the Normal for
luda_lava [24]

Answer:

50 N.

Explanation:

On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:

\displaystyle \begin{aligned} F_N = F_g & = ma \\ & = mg\end{aligned}

The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:

\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}

In conclusion, the normal force acting on the block is 50 N.

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2 years ago
Ganymede is one of the many moons of Jupiter. It is nearly spherical in shape. It is larger than the planet Mercury and slightly
lesya [120]
Moons revolve around a planet, and planets revolve around the sun. Ganymede is considered a moon because it revolves around the planet Jupiter, therefore, it is a moon. :) I hope this helps! 
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3 years ago
Read 2 more answers
Question in pic.. thanks!
11Alexandr11 [23.1K]

That's false. Displacement would be (r2 - r1) .

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3 years ago
A wrestler weighs in for the first match on the moon. will the athlete weigh more or less on the moon than he does on Earth?
Firdavs [7]
The moon<span> is 1/4 the size of </span>Earth<span>, so the </span>moon's<span> gravity is much less than the </span>earth's gravity, 83.3% (or 5/6) less to be exact. Finally, "weight<span>" is a measure of the gravitational pull between two objects. So of course you would </span>weigh<span> much less on the </span>moon<span>.</span>
4 0
3 years ago
Vector A has a magnitude of 50 units and points in the positive x direction. A second vector, B , has a magnitude of 120 units a
Alex Ar [27]

A) Vector A

The x-component of a vector can be found by using the formula

v_x = v cos \theta

where

v is the magnitude of the vector

\theta is the angle between the x-axis and the direction of the vector

- Vector A has a magnitude of 50 units along the positive x-direction, so \theta_A = 0^{\circ}. So its x-component is

A_x = A cos \theta_A = (50) cos 0^{\circ}=50

- Vector B has a magnitude of 120 units and the direction is \theta_B = -70^{\circ} (negative since it is below the x-axis), so the x-component is

B_x = B cos \theta_B = (120) cos (-70^{\circ})=41

So, vector A has the greater x component.

B) Vector B

Instead, the y-component of a vector can be found by using the formula

v_y = v sin \theta

Here we have

- Vector B has a magnitude of 50 units along the positive x-direction, so \theta_A = 0^{\circ}. So its y-component is

A_y = A sin \theta_A = (50) sin 0^{\circ}=0

- Vector B has a magnitude of 120 units and the direction is \theta_B = -70^{\circ}, so the y-component is

B_y = B sin \theta_B = (120) sin (-70^{\circ})=-112.7

where the negative sign means the direction is along negative y:

So, vector B has the greater y component.

8 0
3 years ago
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