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harkovskaia [24]

3 years ago

15

Gravity accelerates any object falling at 9.8 m/s2. What is the force that will be applied to the ground if an 88 kg box falls f

rom the top of a shelf to the ground?

1 answer:

saw5 [17]3 years ago

6 0

Answer:

The force is of the weight <u>862,4 Newtons.</u>

<u></u>

Explanation:

We have the formula:

w = m * g

Data:

w = weight = ¿?

m = mass = 88 kg

g = gravity = 9,8 m/s

Replacing according formula:

w = 88 kg * 9,8 m/s

Resolving:

w = 862,4 N

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What is 74000000 in scientific notation

bekas [8.4K]

It is 7.4x(10 to the power of 7)

3 0

3 years ago

Which describes the position on a horizontal number line

Savatey [412]

Answer:

It is formed by a horizontal number line, called the x-axis, and a vertical number line, called the y-axis.

Explanation:

3 0

3 years ago

Ryan places 0.150 kg of boiling water in a thermos bottle. How many kgs of ice at –12.0 °C must Ryan add to the thermos so that

Greeley [361]

Answer:

The value is $m_i = 0.0234 \ kg$

Explanation:

Generally from the calorimetry principle we have that

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

So here heat gained water is mathematically represented as i.e

$Q_w = m_w * c_w * (T_w - T )$

substituting 0.150 kg for $m_w$ , 4200 J/kg.°C for $c_w$ , 100°C for $T_w$ and 75°C for T

We have

$Q_w = 0.150 * 4200 * (100 - 75 )$

$Q_w =15750 \ J$

The Heat loss by the ice is mathematically represented as

$Q_i = Q_1 + Q_2 + Q_3$

Here $Q_1$ is the energy to move the ice to its melting point which is evaluated as

$Q_1 = m_i * c_i * ( T_o -T_i)$

Here $m_i$ is the mass of ice

$c_i$ is the specific heat of ice with value $2.05 * 10^3 J/kg.°C$

$T_o$ temperature of ice at melting point with value 0°C

$T_i$ is the temperature of ice with value -12°C

$Q_2$ is the energy to move the ice from its its melting point to liquid which is evaluated as

$Q_2 = m_i * L$

Here L is the Latent heat of melting of ice with value $334 * 10^3 J/kg$

$Q_3$ is the energy to move the ice from liquid to the equilibrium temperature which is evaluated as

$Q_1 = m_i * c_w * ( T -T_o)$

So

$Q_i = m_i [ c_i * ( T_o -T_i) + L + c_w * ( T -T_o) ]$

=> $Q_i = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3 + 4200 * ( 75 - 0) ]$

From

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

We have that

$m_i * 673600 =15750$

=> $m_i = \frac{15750}{673600}$

=> $m_i = 0.0234 \ kg$

8 0

4 years ago

In which type of circuit does charge move in only one direction?

Lesechka [4]

C. A combined circuit
That’s the answer

3 0

3 years ago

A diver can change his rotational inertia by drawing his arms andlegs close to his body in the tuck position. After he leaves th

NeX [460]

Answer:

3.14946 rad/s

Explanation:

$I_i$ = Intial moment of inertia

$I_f$ = Final moment of inertia

$\omega_i$ = Initial angular velocity

$\omega_f$ = Final angular velocity = $\dfrac{2}{1.33}\times 2\pi\ rad/s$

$\dfrac{I_f}{I_i}=\dfrac{1}{3}$

In this system the angular momentum is conserved

$L_i=L_f\\\Rightarrow I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_i=\dfrac{I_f\omega_f}{I_i}\\\Rightarrow \omega_i=\dfrac{1\times \dfrac{2}{1.33}\times 2\pi}{3}\\\Rightarrow \omega_i=3.14946\ rad/s$

The angular velocity when the diver left the board is 3.14946 rad/s

3 0

3 years ago