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geniusboy [140]
3 years ago
6

Mary slides down a snow-covered hill on a large piece of cardboard and then slides across a frozen pond at a constant velocity o

f 2.40 m/s. After Mary has reached the bottom of the hill and is sliding across the ice, Sue runs after her at a velocity of 4.40 m/s and hops on the cardboard. How fast do the two of them slide across the ice together on the cardboard? Mary's mass is 69.0 kg and Sue's is 56.0 kg. Ignore the mass of the cardboard and any friction between the cardboard and the snow and/or ice. (Indicate the direction with the sign of your answer.) m/s
Physics
1 answer:
Tresset [83]3 years ago
3 0

Answer: 3.288 m/s

Explanation:

Given

Mass of Mary, m1 = 69 kg

Mass of Sue, m2 = 56 kg

Speed of Mary, v1 = 2.4 m/s

Speed of Sue, v2 = 4.4 m/s

Speed of the 2 of them, v = ?

We solve this using the principle of conservation of linear momentum

m1.v1 + m2.v2 = m1.v + m2.v

m1.v1 + m2.v2 = (m1 + m2) v

v = [(m1.v1) + (m2.v2)] / (m1 + m2)

v = [(69 * 2.4) + (56 * 4.4)] / (69 + 56)

v = (164.6 + 246.4) / 125

v = 411 / 125

v = 3.288 m/s

Thus, the speed at which both Mary and Sue slide together across the ice is 3.288 m/s

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Given Information:

KEa = 9520 eV

KEb = 7060 eV

Electric potential = Va = -55 V

Electric potential = Vb = +27 V

Required Information:

Charge of the particle = q = ?

Answer:

Charge of the particle = +4.8x10⁻¹⁸ C

Explanation:

From the law of conservation of energy, we have

ΔKE = -qΔV

KEb - KEa = -q(Vb - Va)

-q = KEb - KEa/Vb - Va

-q = 7060 - 9520/27 - (-55)

-q = 7060 - 9520/27 + 55

-q = -2460/82

minus sign cancels out

q = 2460/82

Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹

q = 2460(1.60x10⁻¹⁹)/82

q = +4.8x10⁻¹⁸ C

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3 years ago
Which of the following is a form of potential energy? O A. Sound energy O B. Elastic energy O C. Light energy O O O D. Kinetic e
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Answer:

O C. Light energy

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it conducts energy in it and is an energy itself.

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2 years ago
An object on earth weighs 120 N. What is it’s mass ?
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120 N = (mass) x (9.8 m/s²)

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A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liqui
GalinKa [24]

Answer:

n_l = 1.97

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

focal length of lens in liquid is

\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

                =\frac{n_{g} -n_{l}}{n_{l}}  [\frac{1}{(n_{g} - 1) f}

rearrange fron_l

n_l = \frac{n_g f_l}{f_l+f(n_g-1)}

n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}

n_l = 1.97

7 0
3 years ago
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