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2 years ago

Explain the forces that keep a rock balanced on a tiny pedestal. Please help me with this!

1 answer:

8 0

Normal force for the rock because that makes an object stable at its position.
static friction because micro-welts hold its particle on its position so it doesn't change in position by a potential energy. Gravity makes it stay on the ground because its force attraction between an object and the earth.
Hope this helps <span />

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A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle

denpristay [2]

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/ $s^{2}$ = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = $\frac{8.04N}{70.2Kg} = 0.1130801680m/s^2$ .

and velocity of V = 0.1130801680m/s.

at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

4 0

2 years ago

A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, $v_{o} = 20.0 m/s$

Angle made by the ramp, $\theta = 22.0^{\circ}$

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH$

where

H = $dsin22^{\circ} = 5sin22^{\circ}$

g = $9.8 m/s^{2}$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ}}$ = 19.06 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m$

(b) now, considering the coefficient of friction bhetween ramp and the ball, $\mu = 0.150$ :

velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH - \mu gd$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5}$ = 18.7 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m$

6 0

2 years ago

A swimming pool contains x (less than 0.02) grams of chlorine per cubic meter. the pool measures 5 meters by 50 meters and is 2

zubka84 [21]

The solution for this problem would be:(10 - 500x) / (5 - x)
so start by doing:
x(5*50*2) - xV + 5V = 0.02(5*50*2)
500x - xV + 5V = 10
V(5 - x) = 10 - 500x
V = (10 - 500x) / (5 - x)
(V stands for the volume, but leaves us with the expression for x)

3 0

2 years ago

A stone was dropped off a cliff and hit the ground with a speed of 96 ft/s. What is the height of the cliff? (Use 32 ft/s2 for t

nadezda [96]

The initial velocity of the stone is 0 ft/s. Given the initial velocity (Vi), final velocity (Vf), and acceleration due to gravity (g), the distance may be calculated through the equation,
d = ((Vf)² - (Vi)²) / 2g
Substituting the known values,
d = ((96 ft/s)² - 0))/ (2x32.2)
The value of d is 143.10 ft.

4 0

3 years ago

How much heat is required to heat 2 kg of water from 25°C to 40°C?

Dominik [7]

Answer:

126000 J

Explanation:

Applying,

Q = cm(t₂-t₁).................. Equation 1

Where Q = Amount of heat, c = specifc heat capacity of water, m = mass of water, t₁ = Initial temperature, t₂ = Final temperature.

From the question,

Given: m = 2 kg, t₁ = 25°C, t₂ = 40°C

Constant: c = 4200 J/kg.°C

Substitute these value into equation 1

Q = 2×4200(40-25)

Q = 2×4200×15

Q = 126000 J

5 0

3 years ago