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2 years ago

If we want to describe work, we must have

Physics

2 answers:

tresset_1 [31]2 years ago

8 0

The answer is C

Hope that helps!

Good luck :)

PtichkaEL [24]2 years ago

6 0

Answer:

The correct answer is option c.

Explanation:

Work is defined as product of the force applied on object and displacement of the object.the direction of the force and displacement of the object are same in direction.It is measures is Joules (J).

$W=F\times d$

Joules(J)= Newton(N) × meter(m)

W = Work done in Joules

F = force applied in Newtons.

d = displace of the object due to applied force in miters

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van object moving with uniform acceleration has a velocity of 10.0 cm/s in the positive x-direction when its x-coordinate is 3.0

creativ13 [48]

Initial velocity of object vi=10.0 cm/s

initial position fo vector of the object is xi=3.09 cm

Final position of vector xf=-5.00cm

then displacement of object s = xf-xi=-5.00-3.09=-8.09cm

time t=2.55 s

s=vit+1/2at2

-5.00 = 11*2.55+1/2*a2.552

a = (-5.00 - 10*2.55*2)/2.552 = 2.94 cm

Acceleration is 2.94 cm.

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4 0

1 year ago

3) A person catches a ball with a mass of 145 g dropped from a height of 60.0 m above his glove. His hand stops the ball in 0.01

iragen [17]

Answer: 87KN

Explanation: F= m(v-u)/t

V= h/t

V= 60/0.0100

v= 6000m/s

M=145g/1000=0.145kg

F= 0.145*6000/0 .001

F= 87000N

F= 87KN

7 0

2 years ago

The pressure in a container will stay the same if you ——?

elena55 [62]

I think it’s D...not sure tho !

8 0

2 years ago

A geological process Select one: A. is limited to acting on rocks B. shapes and changes the earth C. starts in the outer atmosph

navik [9.2K]

It is B i believe thank and good bye

5 0

2 years ago

A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th

saveliy_v [14]

Answer :

Velocity will be $3.28\times 10^{-11}m/sec$

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere = $=1.6+\frac{1.5}{2}=2.35mm = 0.00235m$

Force on electron is given by $F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}$ , here q is charge on sphere and e is charge on electron

$F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N$

This force work as centripetal force

So $F=\frac{mv^2}{r}$

$4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}$

v = $=0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec$

6 0

2 years ago