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Neporo4naja [7]
2 years ago
13

What is the force when ur pulling heavy furniture that wont budge

Physics
2 answers:
Natali5045456 [20]2 years ago
4 0
I think the answer is friction! Hope it helps!
elena55 [62]2 years ago
3 0

Answer:

Gravity? is it multiple choice?

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Just wanna make sure im right
NikAS [45]
It is right have a good day

explanation step by step
6 0
3 years ago
When light is reflected by a mirror, the angle of incidence is always A. equal to the angle of reflection. B. less than the angl
ankoles [38]
When light is reflected by a mirror, the angle of incidence is always <span>A. equal to the angle of reflection. We know this by the Law of Reflection.</span>
6 0
2 years ago
Read 2 more answers
In the Bohr model of the hydrogen atom, an electron({rm mass};m=9.1; times 10^{ - 31;}{rm kg}) orbits a proton at a distance of
max2010maxim [7]

Answer:

n=6.56×10¹⁵Hz

Explanation:

Given Data

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

To find

Revolutions per second

Solution

Let F be the force of attraction

let n  be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by

F=mω²r......................where ω=2π  n

F=m4π²n²r...............eq(i)

as the values given where

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

we have to find n from eq(i)

n²=F/(m4π²r)

n^{2} =\frac{8.2*10^{-8} }{9.11*10^{-31}* 4\pi^{2} *5.3*10^{-11}  }\\ n^{2}=4.31*10^{31}\\ n=\sqrt{4.31*10^{31}}\\ n=6.56*10^{15}Hz

8 0
3 years ago
What force is needed to accelerate an object 5 m/s if the object has a mass of 10kg?
nevsk [136]
The <span>force that is needed to accelerate an object 5 m/s if the object has a mass of 10kg 50N because you multiply 5 and 10</span>
6 0
3 years ago
A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate
liq [111]

Answer:

3 cm

Explanation:

According to the question,

D=1.5 m.

d=9 cm.

\lambda =0.9 mm.

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

y=2\frac{\lambda (D)}{d}.

Substitute all the variable in above equation.

y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}.

y=3 cm.

5 0
2 years ago
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