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Rudiy27
3 years ago
10

The resultant force acting on an object of mass 5.0kg varies with time as shown. the object is initially at rest.

Physics
1 answer:
il63 [147K]3 years ago
8 0
<span>particle varies with time as shown in the diagram. ... resultant has a magnitude equal to 8.0. .... A constant force F is applied to a body of mass m that initially is headed east at velocity .... If the resultant force acting on a 2.0-kg object is equal to ..... A ball of mass mB is released from rest and acquires velocity of magnitude vB ...</span><span>
</span>
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How does the mass of an object affect its acceleration?
solong [7]

Answer:

as the mass of an object is increased, the acceleration of the object is decreased & vice-versa.

Explanation:

hope this helps!

3 0
2 years ago
A man has a power of 90 W and mass 60 kg runs up a staircase in 40 s. If each staircase is 20cm high find the number of steps? (
Svetlanka [38]
I depends on three area of the staircase
4 0
3 years ago
What is the name of the atomic model in which electrons are treated as waves
chubhunter [2.5K]

Answer:

The atomic model in which electrons are treated as waves is called the wave mechanical model of the atom or the quantum mechanical model of the atom. Principal energy levels contains energy sublevels

5 0
3 years ago
A closely wound, circular coil with radius 2.50 cmcm has 740 turns. Part A What must the current in the coil be if the magnetic
Vika [28.1K]

Answer:

The current in the coil is 4.086 A

Explanation:

Given;

radius of the circular coil, R = 2.5 cm = 0.025 m

number of turns of the circular coil, N = 740 turns

magnetic field at the center of the coil, B = 0.076 T

The magnetic field at the center of the coil is given by;

B = \frac{N\mu_o I}{2R}

where;

μ₀ is permeability of free space = 4 x 10⁻⁷ m/A

I is the current in the coil

R is radius of the coil

N is the number of turns of the coil

The current in the circular coil is given by

B = \frac{N\mu_o I}{2R} \\\\I = \frac{2BR}{N\mu_o} \\\\I =\frac{2*0.076*0.025}{740*4\pi*10^{-7}} \\\\I = 4.086 \ A

Therefore, the current in the coil is 4.086 A

3 0
3 years ago
A ball is launched from ground level at 20 m/s at an angle of 40° above the
DedPeter [7]

(a) The ball's height <em>y</em> at time <em>t</em> is given by

<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :

0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )

<em>t</em> = 0   or   (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 <em>g t</em>

<em>t</em> = (40 m/s) sin(40º) / <em>g</em>

<em>t</em> ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So

0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>

where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :

<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)

<em>y</em> ≈ 8.4 m

8 0
2 years ago
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