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Lelechka [254]
3 years ago
15

Suppose you increase your walking speed from 7 m/s to 11 m/s in a period of 1 s. What is your acceleration?

Physics
1 answer:
QveST [7]3 years ago
6 0
Your acceleration would be four
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The core of a star must be at temperature of _____ degrees Celsius for hydrogen fusion to take place. 10,000 100,000 1,000,000 1
disa [49]
The core of a star must be at the temperature of 10,000,000 degrees Celsius for hydrogen fusion to begin. 
6 0
3 years ago
A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
Setler [38]

Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

   work done by the gravity = ?          

   work done by the tension = ?            

Work done by the gravity = - m g Δh            

 Δ h = 2 + 2 = 4 m                                                                

Work done by the gravity =- 3 \times 9.8 \times 4

                                           = -117.6 J                  

work done by gravity is equal to -117.6 J            

Work done by tension will be equal to zero.        

Zero because tension is always perpendicular to velocity

work done by tension is equal to 0 J                          

7 0
3 years ago
The typical unit for a period used with Kepler's third law is
ollegr [7]
Well, if you're using the law to work with periods of Earth satellites,
then the most convenient unit is going to be 'hours' for the largest
orbits, or 'minutes' for the LEOs.

But if you're using it to work with periods of planets, asteroids, or
comets, then you'd be working in days or years.
6 0
3 years ago
A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

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

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0
3 years ago
As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax
AysviL [449]

Answer:

I=2.766\ kg.m^2

Explanation:

We have:

diameter of the wheel, d=0.88\ m

weight of the wheel, w_w=280\ N

mass of hanging object to the wheel, m_o=6.32\ kg

speed of the hanging mass after the descend, v_o=4\ m.s^{-1}

height of descend, h=2.5\ m

(a)

moment of inertia of wheel about its central axis:

I=\frac{1}{2} m.r^2

I=\frac{1}{2} \frac{w_w}{g}.r^2

I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2

I=2.766\ kg.m^2

3 0
3 years ago
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