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3 years ago

6

A football player kicks a ball horizontally off a hill with an initial velocity of 42.0 m/s. It travels a horizontal distance of

59.2m. How y'all was the hill he kicked the ball off of?

1 answer:

damaskus [11]3 years ago

7 0

Explanation:

.........................

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koban [17]

Answer:

It's C

Explanation:

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2 years ago

The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°c _____.

Fiesta28 [93]

Answer:

4.2 J

Explanation:

Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K

From specific heat capacity,

Q = cmΔt.............................. Equation 1

Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.

Given: m = 1 g = 0.001 kg, Δt = 1 °C

Constant : c = 4200 J/kg.°C

Substitute into equation 1

Q = 0.001×4200(1)

Q = 4.2 J.

Hence the energy absorbed or lost = 4.2 J

6 0

3 years ago

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

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3 years ago

EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE AHHHHHHHHHHHHHHHHHHHHHHHHHHH COCOA PEBBLES RULE HEHEHEHEHEHEHE

Andrew [12]

Answer:

AAAAAAAAHHHDHSFJSBDKFANEDASNXDEHAHDBWEEEEHJDBHBAH

AAHHH YES THEY DOO AAAHHH EEHSFJWEHASAAAAHJEHER

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2 years ago

Read 2 more answers

I am struggling on this physics question. Brainly is my last hope. Could somebody please provide an answer to this question, wit

Aleks [24]

1) 29.4 N

The force of gravity between two objects is given by:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

M and m are the masses of the two objects

r is the separation between the centres of mass of the two objects

In this problem, we have

$M=5.97\cdot 10^{24} kg$ (mass of the Earth)

$m=3.0 kg$ (mass of the box)

$r=R=6.37\cdot 10^6 m$ (Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)

Substituting into the equation, we find F:

$F=\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24})(3.0)}{(6.37\cdot 10^6)^2}=29.4 N$

2) $g=9.8 m/s^2$

Let's now calculate the ratio F/m. We have:

F = 29.4 N

m = 3.0 kg

Subsituting, we find

$\frac{F}{m}=\frac{29.4}{3.0}=9.8 N/kg = 9.8 m/s^2$

This is called acceleration of gravity, and it is the acceleration at which every object falls near the Earth's surface. It is indicated with the symbol $g$ .

We can prove that this is the acceleration of the object: in fact, according to Newton's second law,

$F=ma$

where a is the acceleration of the object. Re-arranging,

$a=\frac{F}{m}$

which is exactly equal to the quantity we have calculated above.

5 0

2 years ago