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Nady [450]
3 years ago
12

What is the minimum work needed to push a 920-kg car 310 m up along a 6.5 ∘incline? Ignore friction.Express your answer using tw

o significant figures.
Physics
1 answer:
AVprozaik [17]3 years ago
7 0

Answer:

2800000J

Explanation:

Parameters given:

Mass = 920kg, weight = 920 * 9.8 = 9016N

Distance = 310m

Angle of inclination = 6.5°

Work done is given as :

W = F*d*cosA

Where A = angle of inclination

W = (9016 * 310 * cos6.5)

W = 2776993.59J

In 2 significant figures, W = 2800000J

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Calculate the density of the football. Use the formula D = m/V where D is the density, m is the mass, and V is the volume. Recor
s344n2d4d5 [400]

Answer:

Detailed explanation:

Density of water=1000kg/m³

Hence mass of water displaced is:

m=d×v

=1000kg/m³×(4.3×10^-3)m³ (volume of water displaced converted to L)

=4.3 kg of water

Hence, mass of football is also 4.3 kg(Archimedes principle)

Thus density of football

=mass÷volume

substitute the mass and volume and solve.

hope this helps

4 0
3 years ago
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. A student thinks that the more protons an atom has, the more reactive it will be. Which will best disprove this student's idea
shepuryov [24]
The best answer to the question that is being stated above would be the first choice. To disprove the claims of the student that you need more protons to have more reactivity, then you proceed to compare the reactivities of lithium (Li) and krypton (Kr). Krypton has more protons than Lithium, but it is less reactive because it is a noble gas.
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3 years ago
In the standing waves experiment, the string has a mass of 31.2 g and a length of 0.7 m. The string is connected to a mechanical
mestny [16]

Answer:

linear density of the string = 4.46 × 10⁻⁴ kg/m

Explanation:

given,

mass of the string = 31.2 g

length of string = 0.7 m

linear density of the string = \dfrac{mass\ of\ string}{length}

linear density of the string = \dfrac{31.2\times 10^{-3}\ kg}{0.7\ m}

linear density of the string = 44.57 × 10⁻³ kg/m

linear density of the string = 4.46 × 10⁻⁴ kg/m

7 0
3 years ago
[O.04H]The table below shows the use of some energy production methods over time.
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3 years ago
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The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo
a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s

The tire angular velocity before stopping is:

\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s

Also its angular decceleration:

\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2

Using the following equation motion we can findout the angle it makes during the deceleration:

\omega^2 - \omega_0^2 = 2\alpha\Delta \theta

where \omega = 0 m/s is the final angular velocity of the car when it stops, \omega_0 = 114rad/s is the initial angular velocity of the car \alpha = 14.75 rad/s2 is the deceleration of the can, and \Delta \theta is the angular distance traveled, which we care looking for:

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\Delta \theta = 440rad or 440/2π = 70 revelutions

4 0
3 years ago
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