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3 years ago

A 71.80 kg person holding a steel ball stands motionless on a frozen lake.

Physics

2 answers:

Svet_ta [14]3 years ago

8 0

Answer:

The answer is A. 3.08 kg

Mars2501 [29]3 years ago

6 0

Answer: 3.08 !! <3

Explanation: im sorry but im not sure y i just got the answer wrong and this is wut it told me was right :)

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9. If everything in the universe – including atoms and particles stop moving, does time stop? Or does time continue even if ever

Vedmedyk [2.9K]

Time stops everything is made out of atoms so if atoms freeze everything freezes

5 0

3 years ago

4. A substance has a density of 0.79 g/cm'. It is soluble in water. List all the possibilities of what it might be How could you

Natasha_Volkova [10]

Answer:

See explanation

Explanation:

Given that the density of the unknown substance is 0.79 g/cm3 and is soluble in water, the possible substances it could be are;

i) t-butanol

ii) ethanol

iii) 2-propanol

iv) acetone

However, the actual identity of the unknown substance can be obtained by carrying out a boiling point test. The four substances listed above have different boiling points. Hence the boiling point of the unknown substance ultimately discloses its identity.

5 0

3 years ago

The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

3 years ago

What is the minimum speed must a salmon jumping at an angle of 35.2 leave the water in m/s?

yuradex [85]

Height of the waterfall is 0.449 m

its horizontal distance will be 2.1 m

now let say his speed is v with which he jumped out so here the two components of his velocity will be

$v_x = vcos35.2 = 0.817 v$

$v_y = vsin35.2 = 0.576 v$

here the acceleration due to gravity is 9.81 m/s^2 downwards

now we can find the time to reach the other end by y direction displacement equation

$\Delta y = v_y * t + \frac{1}{2} at^2$

$-0.449 = 0.576 * v *t - \frac{1}{2}*9.81 * t^2$

also from x direction we can say

$\Delta x = v_x * t$

$2.1 = 0.817 v* t$

now we have

$v* t = 2.57$

we will plug in this value into first equation

$- 0.449 = 0.576 * 2.57 - 4.905 * t^2$

$1.93 = 4.905 * t^2$

$t = 0.63 s$

now as we know that

$v* t = 2.57$

t = 0.63 s

$v = \frac{2.57}{0.63}$

$v = 4.1 m/s$

so his minimum speed of jump is 4.1 m/s

8 0

3 years ago

Students were discussing a problem in which the class was asked to find the acceleration of a cart rolling up and down an inclin

velikii [3]

Sasha is correct: No, the velocity is changing at the top so the acceleration cant be zero.

Explanation:

The acceleration of an object is equal to the rate of change in velocity of the object:

$a=\frac{\Delta v}{\Delta t}$ (1)

where

$\Delta v$ is the change in velocity

$\Delta t$ is the time interval

For the cart on the ramp, as the cart reaches the top of the ramp, its velocity becomes temporarily zero:

v = 0

This is the origin of Malia's mistake.

However, this only lasts a moment; in reality, its velocity is changing, in particular its direction is changing (from upward to downward).

As we can see from eq.(1), a non-zero change in velocity implies a non-zero acceleration. Therefore, Sasha is right, since the cart has a non-zero acceleration.

Learn more about acceleration:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

4 0

3 years ago