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3 years ago

Please help ASAP!!

1 answer:

6 0

There is kinetic energy when it is sitting at the top, then as it goes towards the bottom, the kinetic energy is transformed into potential energy.

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A ball at the end of a string is swinging in a horizontal circle of radius 0.85 m. The ball makes exactly 3 revolutions per seco

Anettt [7]

Answer:

It's centripetal acceleration is 301.7 m/s²

Explanation:

The formula to be used here is that of the centripetal acceleration which is

ac = rω²

where ac is the centripetal acceleration = ?

ω is the angular velocity = 3 revolutions per second is to be converted to radian per second: 3 × 2π = 3 × 2 × 3.14 = 18.84 rad/s

r is the radius = 0.85 m

ac = 0.85 × 18.84²

ac = 301.7 m/s²

It's centripetal acceleration is 301.7 m/s²

8 0

2 years ago

Suppose that you are standing on a train accelerating at 0.20g (where g is the acceleration due to gravity). What minimum coeffi

prisoha [69]

Answer:

0.2

Explanation:

The given parameters are;

The acceleration of the train, a = 0.2·g

The mass of the person standing on the train = m

Let μ represent the coefficient of static friction, we have;

The force acting on the person, F = m × a = m × 0.2·g

The force of friction acting between the feet and the floor, $F_f$ = m·g·μ

For the person not to slide we have;

The force acting on the person = The force of friction acting between the feet and the floor

F = $F_f$

∴ m × 0.2·g = m·g·μ

From which we get;

0.2 = μ

The coefficient of static friction that must exist between the feet and the floor if the person is not to slide, μ = 0.2.

7 0

2 years ago

A block of aluminum at a temperature of T1 = 32 degrees C has a mass of m1 = 12 kg. It is brought into contact with another bloc

Mariulka [41]

Answer: $T=21.33 ^{\circ}$

Explanation:

Given

mass of First Block $m_1=12 kg$

Temperature $T_1=32^{\circ}$

mass of second block $m_2=0.5 m_1=6 kg$

Temperature $T_2=9^{\circ}$

Heat capacity of aluminium c=899 J/kg-K

Final Temperature acquired by both blocks at steady state

Heat loss first block =Heat gain by second block

$m_1\times c\times (32-T)=m_2\times c\times (T-9)$

$12\times 899\times (32-T)=6\times 899\times (T-9)$

$2\times 32=3T$

$T=\frac{64}{3}=21.33 ^{\circ}$

5 0

2 years ago

How fast can Usain Bolt run if it takes him 9.9 s to run 100m?

Arte-miy333 [17]

Answer:

27.8 mph

Explanation:

May I have brainliest please? :)

8 0

2 years ago

Read 2 more answers

What mass of ice (in g) can be melted if 27.2 kJ of thermal energy are added at the freezing point? Use molar mass = 18.02 g/mol

san4es73 [151]

Answer : The mass of ice melted can be, 3.98 grams.

Explanation :

First we have to calculate the moles of ice.

$Q=\frac{\Delta H}{n}$

where,

Q = energy absorbed = 27.2 kJ

$\Delta H$ = enthalpy of fusion of ice = 6.01 kJ/mol

n = moles = ?

Now put all the given values in the above expression, we get:

$27.2kJ=\frac{6.01kJ/mol}{n}$

$n=0.221mol$

Now we have to calculate the mass of ice.

$\text{Mass of ice}=\text{Moles of ice}\times \text{Molar mass of ice}$

Molar mass of ice = 18.02 g/mol

$\text{Mass of ice}=0.221mol\times 18.02g/mol=3.98g$

Thus, the mass of ice melted can be, 3.98 grams.

3 0

3 years ago