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3 years ago

How you can speed up the dissolving process when preparing juice from frozen concentrate

Physics

1 answer:

Katyanochek1 [597]3 years ago

8 0

If you stir the juice it increases the surface area.

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A man has a mass of 70kg. Calculate his weight on earth where the gravitational strength is 10 N/kg

maw [93]

Answer:

So the weight of the body is 700N

Explanation:

mass of the object=70kg

force of gravity=10N/kg

weight of the object=w-?

as we know that

weight=mass × force of gravity

weight=70kg × 10N/kg

WEIGHT OF THE BODY=700N

I HOPE IT WILL HELP YOU

GOOD LUCK FOR THE ASSIGNMENT

4 0

4 years ago

Two identical spheres with m=3 kg each are rigidly attached to the rotating light rigid structure. The velocity of the spheres i

Dennis_Churaev [7]

Answer:

Angular momentum of the system is 4.8 kg m^2/s

Explanation:

As we know that the angular momentum of the system of masses is given as

$L = 2I\omega$

here we have

$I = m(0.2 + 0.4)^2$

$I = 3(0.6)^2$

now we also know that

$\omega = \frac{v}{0.2 + 0.4}$

$0.6 \omega = 2 m/s$

now we have

$L = 2(2)(0.6)(2)$

$L = 4.8 kg m^2/s$

3 0

3 years ago

Energy of a SpacecraftVery far from earth(at R=\infty), a spacecraft has run out of fuel and its kineticenergy is zero. If only

Firdavs [7]

Answer:

$s_e=\sqrt{\frac{2GM_e}{R_e^2}}$

Explanation:

In this case mechanical energy is conserved, which means that the sum of the initial kinetic energy and initial potential gravitational energy will be equal to the sum of the final kinetic energy and final potential gravitational energy:

$K_i+U_i=K_f+U_f$

Which in our case will be:

$\frac{mv_i^2}{2}+\frac{-GM_em}{r_i^2}=\frac{mv_f^2}{2}+\frac{-GM_em}{r_f^2}$

Which, since $v_i=0m/s$ , $r_i=infinity$ , $r_f=R_e$ , $v_f=s_e$ and canceling <em>m</em> means that:

$\frac{s_f^2}{2}=\frac{GM_e}{R_e^2}$

Solving for the final velocity we get:

$s_e=\sqrt{\frac{2GM_e}{R_e^2}}$

6 0

3 years ago

Help me frfr I don’t understand

Art [367]

The points are

(1,10)

(6,0)

$\boxed{\sf slope(m)=\dfrac{y_2-y_1}{x_2-x_1}}$

$\\ \sf\longmapsto m=\dfrac{0-10}{6-1}$

$\\ \sf\longmapsto m=\dfrac{-10}{5}$

$\\ \sf\longmapsto m=-2$

6 0

3 years ago

A body of mass 2.78 kg is pushed straight upward by a 31.3 N vertical force. What is its acceleration (in m/s2)

Bingel [31]

To solve this problem we will apply the concepts related to Newton's second law for which the product of mass and acceleration is defined as the force applied to an object. Mathematically this is,

$F_{net} = ma \rightarrow a = \frac{F_{net}}{m}$