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MrRissso [65]
3 years ago
5

A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H

ELP
Physics
1 answer:
Aleks04 [339]3 years ago
3 0

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .

Explanation:

Dont report my answer please

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Answer: biotic

Explanation:

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3 years ago
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
3 years ago
A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an
Andreas93 [3]

Answer:

170 N

Explanation:

Given in the question that, work a bulldozer can do = 4500 J

<h3>Step 1</h3>

We will use trigonometry identity to find the distance bulldozer will travel up the hill

sin(35) = opp/hypo

sin(35) = 15/hypo

hypo = 15/sin(35)

hypo = 26.15m

<h3>Step 2</h3>

Formula to use

work done = force × distance

Plug values in the above formula

4500 = force x 26.15

force = 4500/26.15

force = 172.08

force ≈ 170 N

<h3 /><h3 /><h3 />

3 0
3 years ago
Read 2 more answers
Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the
Natasha_Volkova [10]

The force per unit length between the two wires is 6.0\cdot 10^{-5} N/m

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

where

\mu_0 = 4\pi \cdot 10^{-7} Tm/A is the vacuum permeability

I_1, I_2 are the currents in the two wires

r is the separation between the two wires

For the wires in this problem, we have

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I_2 = 3.45 A

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Substituting into the equation, we find

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m

Learn more about current and magnetic fields:

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4 0
3 years ago
Calculate the force exerted on the wall assuming that force is horizontal and using the data in the schematic representation of
Jlenok [28]

Answer:

1.93 x 10∧3 N

Explanation:

The picture attached shows the calculation

8 0
3 years ago
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