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3 years ago

An ice skater's mass is 5 Kg, she hits a wall with a force of 10 N. What was her acceleration?

Physics

2 answers:

Nataliya [291]3 years ago

6 0

It’s Prince sunghoon on the iceeee

Harman [31]3 years ago

3 0

Answer:

What was her acceleration and distance she moved during her ... Identify the following as a scalar or vector: the mass of an object, the number of leaves on a ... Once the chair is in motion, a 127 N horizontal force keeps it moving at a constant velocity. a. ... A 65 kg skater standing on frictionless ice throws a 0.15 kg snowball ...

Explanation:

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5. the combustion of a single molecule of methane produces about 10 ev of energy. a methane molecule has a mass of 16 amu. the f

Alchen [17]

The mass of methane will produce as much energy as a single gram of uranium-235 1,287 kilograms of methane. Option D is correct.

<h3>What is uranium?</h3>

Uranium, with the atomic number 92 and the symbol U in the periodic table, is a weakly radioactive element. One of the heavy metals that may be used as a concentrated energy source is uranium.

Given data;

⇒One mole of U-235 = 235 grams

=> 1 gram of U-235 = 1/235 moles

⇒1 mole of U-235 = 6.023 x 10²³ atoms

The no of atoms is;

=> 1/235 moles of U-235

⇒N = 6.023 x 10²³/235 atoms

⇒N=25.6 x 10²⁰ atoms

If one atom of U-235 gives is 189 x 10⁶ eV energy,25.6 x 10²⁰ atoms of U-235 gives;

=> 25.6 x 10²⁰ atoms of U-235 gives;

E = 25.6 x 10²⁰ x 189 x 10⁶

E= 4.8 x 10²⁹ eV energy

One methane molecule produces;

E = 10 eV of energy

=> To produce 4.8 x 10²⁹ eV energy, no. of molecules required;

= 4.8 x 10²⁸

6.032 x 10²³ molecules of methane = 16 gms

=> 4.8 x 10²⁸ molecules of methane = 16 x 4.8 x 10²⁸/ 6.032 x 10²³ gms

m= 12.75 x 10⁵ gms

m= 1275 kilograms of methane

m≅ 1287 kilograms of methane

1 gram of U-235 has the same amount of energy as 1287 kilograms of methane.

Hence option D is correct,

To learn more about uranium refer to the link;

brainly.com/question/9099776

#SPJ1

7 0

2 years ago

A concert loudspeaker suspended high off the ground emits 28.0 W of sound power. A small microphone with a 0.700 cm2 area is 55.

grandymaker [24]

Explanation:

It is known that wave intensity is the power to area ratio.

Mathematically, I = $\frac{P}{A}$

As it is given that power is 28.0 W and area is $7 \times 10^{-5} m^{2}$ .

Therefore, sound intensity will be calculated as follows.

I = $\frac{P}{A}$

= $\frac{28.0 W}{4 \times 3.14 \times 7 \times 10^{-5} m^{2}}$

= $0.318 \times 10^{5} W/m^{2}$

or, = $3.18 \times 10^{4} W/m^{2}$

Thus, we can conclude that sound intensity at the position of the microphone is $3.18 \times 10^{4} W/m^{2}$ .

7 0

3 years ago

When 1 kg of water and 1 kg of wood absorb the same amount of heat, the change in temperature of the wood is greater than the ch

vampirchik [111]

Your Answer Will Be C

8 0

3 years ago

A schoolbus accelerates to 65 mph and enters the freeway. It travels for 2.3 hours at that speed while on the freeway. What's th

PtichkaEL [24]

Answer:

The distance travelled on the freeway is 149.5 miles.

Explanation:

The school bus travels on the freeway at constant speed. According to the statement, we need to calculate the distance travelled by the vehicle by means of the following formula:

$x = v\cdot t$ (1)

Where:

$x$ - Traveled distance, in miles.

$v$ - Speed, in miles per hour.

$t$ - Time, in hours.

If we know that $v = 65\,\frac{mi}{h}$ and $t = 2.3\,h$ , then the distance travelled by the school bus is:

$x = v\cdot t$

$x = \left(65\,\frac{mi}{h} \right)\cdot (2.3\,h)$

$x = 149.5\,mi$

The distance travelled on the freeway is 149.5 miles.

5 0

3 years ago

Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th

lapo4ka [179]

Answer:

$T_f=5.0116^{\circ}C$

Explanation:

Given:

mass of water, $m_w=0.25\ kg$

initial temperature of water, $T_i_w=20^{\circ}C$

initial temperature of pan, $T_i_p=173^{\circ}C$

mass of pan, $m_p=0.6\ kg$

mass of water evapourated, $m_v=0.03\ kg$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

specific heat of aluminium pan, $c_a=900\ J.kg^{-1}.K^{-1}$

latent heat of vapourization, $L=2256000\ J.kg^{-1}$

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

$m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})$

$0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)$

$T_f=5.0116^{\circ}C$

5 0

3 years ago