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4 years ago

14

While an object is moving at a constant 20m/s a 5 N force pushes the object to the left at the same time 5 N force is pushing th

e object to the right what will the objects velocity be after 10 seconds

1 answer:

KatRina [158]4 years ago

5 0

0N because the two forces are canceling each other out

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In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let's find out by doing a

MaRussiya [10]

Answer:

$v= \frac{10\times10^{-3} V}{70}$ m/s

Explanation:

Assumption: bullet leaves the muzzle at a speed of V m/s

and velocity of push received by the man be v m/s

According to newton's third law to every action there is always an equal and opposite reaction.

therefore,

mass of man× velocity = mass of bullet×its velocity

⇒70×v= 10×10^-3 ×V

solving the above eqaution we get

therefore $v= \frac{10\times10^{-3} V}{70}$ m/s

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4 years ago

At what temperature do the fahrenheit and kelvin scales have the same numerical value?

Anettt [7]

Answer: 574.59 K

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4 years ago

Why is work measured in

erastova [34]

Answer:

C) the equation consists of two factors:

work = force * distance

W = F · S

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2 years ago

A small convex mirror is placed 60 cm from the pole and on the axis of a large concave mirror of radius of curvature 200 cm. The

umka2103 [35]

Answer:

Calculate 'R' of convex mirror and height of the real image

the radius of the convex mirror is 48cm

Explanation:

Distance between convex and concave mirror is =60cm

Radius of the concave mirror (R) = 200cm

For the concave mirror, u = ∞

V = {R}/{2}=100cm

The object for the convex mirror and the final image is on the pole of the concave mirror, and distance between convex and concave mirror is 60cm

u_1=60-100 =-40cm

Object will be behind the convex mirror

1/f=1/40+1/60

f=24cms

the radius of the convex mirror is 48cm

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3 years ago

An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that

zzz [600]

Answer with Explanation:

We are given that

Charge on alpha particle=q=2 e= $2\times 1.6\times 10^{-19} C$

1 e= $1.6\times 10^{-19} C$

Mass of alpha particle=m= $6.64\times 10^{-27} kg$

Potential difference,V= $1.97\times 10^6 V$

Magnetic field,B=3.49 T

a.Speed of alpha particle=v= $\sqrt{\frac{2 qV}{m}}$

By using the formula

$v=\sqrt{\frac{2\times 2\times 1.6\times 10^{-19}\times 1.97\times 10^6}{6.64\times 10^{-27}}$

$v=1.38\times 10^7 m/s$

b.Magnetic force,F $=qvB=2\times 1.6\times 10^{-19}\times 1.38\times 10^7\times 3.49=1.5\times 10^{-11} N$

F= $1.5\times 10^{-11} N$

c.Radius of circular path, r= $\frac{mv^2}{F}$

$r= \frac{6.64\times 10^{-27}\times (1.38\times 10^7)^2}{1.5\times 10^{-11}}$

r=0.084 m

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3 years ago