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3 years ago

10

Two ice skaters stand facing each other at rest on a frozen pond. They push off against one another and the 48 kg skater acquire

s a speed of 0.67 m/s. If the other skater acquires a speed of 0.83 m/s, what is her mass in kilograms?

2 answers:

BaLLatris [955]3 years ago

5 0

Answer:

The mass from the second skater is 40.2Kg

Explanation:

To solve this problem it is necessary to apply the concepts related to conservation of the moment.

By definition for the particular case we have to,

$m_1u_1 = m_2v_2$

Where,

$m_1$ = Mass first skater

$u_1$ = Speed of first skater

$v_2$ = Velocity of second skater

Our values are given as,

$m_1= 48Kgu_1 = 0.67m/sv_2 = 0.8m/s$

Replacing in our equation and re-arrange to find $m_2,$

$m_2 = \frac{m_1u_1}{v_2}m_2 = \frac{(48)(0.67)}{0.8}m_2 = 40.2Kg$

The mass from the second skater is 40.2Kg

Papessa [141]3 years ago

5 0

Answer:

38.75 kg

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = mv+m'v'................... Equation 1

Note: Since both skater were initially at rest, u = u' = 0 m/s.

Also Note that both skater moves in opposite direction after collision, and let the direction of the first skater be positive and that of the second skater be negative.

mv = -m'v'.......................... Equation 2

make m' the subject of the equation above,

m' = -mv/v'..................... Equation 3

Given: m =48 kg, v = 0.67 m/s, v = -0.83 m/s (opposite direction to the first skater)

substitute into equation 3

m' = -48(0.67)/-0.83

m' = -32.16/-0.83

m' = 38.75 kg

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I need help understanding this concept . Would appreciate it so much. Thank you

Firlakuza [10]

Answer:

Explanation:

Both these questions are based on the Universal Law of Gravitation, which is given by :

F = Gm1m2 / r²

2) F = 6.67 x 10⁻¹¹ x 8 x 10³ x 1.5 x 10³ / 1.5 x 1.5

F = 6.67 x 10⁻⁵ x 8 / 1.5

F = 35.57 x 10⁻⁵ N

3) F = 6.67 x 10⁻¹¹ x 7.5 x 10⁵ x 9.2 x 10⁷ / 7.29 x 10⁴

F = 6.67 x 10⁻³ x 7.5 x 9.2 / 7.29

F = 63.13 x 10⁻³ N

7 0

3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago

What is the total resistance in the circuit? (include unit in answer - ohms)

solmaris [256]

Answer:60 ohms

Explanation:

R1=30 ohms

R2=15 ohms

R3=15 ohms

Let the total resistance be R

R=R1 + R2 + R3

R=30 + 15 +15

R=60

Total resistance is 60 ohms

4 0

3 years ago

What is the equation used to calculate the total amount of energy used by an appliance?

charle [14.2K]

Answer:

Power = Current × Voltage

Explanation:

Units:

Power = Watts

Current = Àmperes

Voltage = Volts

8 0

3 years ago

PRACTICE ANOTHER A cube of wood having an edge dimension of 19.7 cm and a density of 647 kg/m3 floats on water. (a) What is the

dedylja [7]

CHECK COMPLETE QUESTION

The cube of wood having an edge dimension of 19.7cm and a density of 647kg/m3 floats on water

(a) What is the distance from the horizontal top surface of the cube to the water level answer in cm

(b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface answer in kg

Answer:

a)6.29cm

b)2.78 kg

Explanation:

Given:

Let us calculate the volume first, we were given dimension as 19.7cm=0.197m

Volume is (0.197 meters)³ = 0.00764m³

Then we can calculate the mass as;

Given mass is 647 kg/m³ x 0.00774m³ = 4.947kg

The weight = mass × acceleration due to gravity

weight = 4.947 x 9.8 N/kg = 48.44N

By Floating we can say the the buoyancy force has to equal the weight (48.44 N) which has

which is equal to the weight of volume of the displaced water. Or the mass, the calculation is the same.

We know that density of fresh water at 20ºC is 998 kg/m³

Then we can calculate the volume of displaced water as

4.947 kg / 998 kg/m³ = 0.00496 m³

We know that the displaced water has a shape of a rectangular solid with 0.197 meters on the two horizontal dimensions, and h as the height to the surface then

V = 0.197²h = 0.00496

0.00496= 0.197²h

h = 0.1278 meters or 12.78 cm

Then the the distance exposed, would be 19.7–12.78 = 6.29 cm

b) ifthe cube is fully submerged, the volume of the displaced water is 0.00774m³

mass of displaced water is 0.00774m³ x 998 kg/m³ = 7.724 kg

Added mass is the mass of the displaced water – mass of block

= 7.724–4.947 = 2.78 kg

5 0

3 years ago