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Lubov Fominskaja [6]
3 years ago
9

PLEASE HELP FAST!!!

Physics
1 answer:
ELEN [110]3 years ago
7 0

Answer:

1.It improved by using the Mars survey probe, that took a portrait of the planet that helped us see the climate of Mars.

Explanation:

I watched the video.

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A vertical spring stretches 9.6 cm when a 1.3 kg block is hung from its end. (a) Calculate the spring constant. This block is th
tatiyna

Answer:

F = - K x

a) K = 1.3 kg * 9.8 m/s^2 / .096 m = 133 kg/sec^2

b)  ω = (K/m)^1/2     angular frequency of SHM

ω = (133 / 1.3)^1/2 = 10.1 / sec

f = 2 π ω = 6.28 * 10.1 / sec = 63.5 / sec

P = 1/f = .0157 sec

5 0
3 years ago
Nuclear sizes are expressed in a unit named
o-na [289]

Answer:

Answer is A) Fermi

Explanation:

Fermi is the expressive unit for nuclear sizes. Fermi = 10^-15 meter.

4 0
3 years ago
A thin walled spherical shell is rolling on a surface. What
ExtremeBDS [4]

Answer:

=\frac{1/3}{5/6} = 0.4

Explanation:

Moment of inertia of given shell= \frac{2}{3} MR^2

where

M represent sphere mass

R -sphere radius

we know linear speed is given as v = r\omega

translational K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2

rotational K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2

total kinetic energy will be

K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2

K.E =\frac{5}{6} MR^2 \omega^2

fraction of rotaional to total K.E

=\frac{1/3}{5/6} = 0.4

8 0
3 years ago
If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0
Ilya [14]
Work of the force = 10 N

Time required for the work = 50 sec

Height = 7 m

We are given with the value of work and time in the question.

Substitute the values in the formula of power and then you'll get the power required.

We know that,

w = Work

p = Power

t = Time

By the formula,

Given that,
Work (w) = 7 m = 70 Joules
Time (t) = 50 sec
Substituting their values,

p = 70/50

p = 1.4 watts

Therefore, the power required is 1.4 watts.

Hope it helps!
3 0
3 years ago
A tennis player swings her 1000 g racket with a speed of 10 m/s. She hits a 60 g tennis ball that was approaching her at a speed
Mashutka [201]

(a)

consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.

m = mass of the tennis ball = 60 g = 0.060 kg

v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s

v = final velocity of the tennis ball after being hit by racket = - 39 m/s

ΔP = change in momentum of the ball

change in momentum of the ball is given as

ΔP = m (v - v₀)

inserting the above values

ΔP = (0.060) (- 39 - 20)

ΔP = - 3.54 kgm/s

hence , magnitude of change in momentum : 3.54 kgm/s

7 0
3 years ago
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