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Hunter-Best [27]
2 years ago
10

Assuming no friction, what is the minimum work needed to push a 1000 kg car 45.0 m up a 12.5 incline?

Physics
2 answers:
Dvinal [7]2 years ago
6 0
Work=Fdcos(theta) so (10000)(45.0)cos(12.5 degrees) = 440000 J
bulgar [2K]2 years ago
4 0

Answer:

92,259.73 Joules

Explanation:

α=12.5°

m=1000 kg

d= 45m

F=m*g*Sin(\alpha)\\°)F= 1000kg*9.81 \frac{m}{s} *Sin (12.5 )\\

F= 1000 kg * 9.81 \frac{m}{s} * 0.2164\\

F= 2,123.27 N\\

W=F*d*Cos (\alpha)\\

W= 2,123.27 N * 45m * Cos (12,5)\\

W= 92,259.73   Joules

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What is the electric force acting between two charges of -0.0045 C and -0.0025 C that are 0.0060 m apart? Use Fe=kq1q2/r^2 and k
oksano4ka [1.4K]

Answer:

D. 2.8 × 10⁹ N

Explanation:

The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.

Fe= k Q₁Q₂/r²

Q₁= -0.0045 C

Q₂= -0.0025 C

r= 0.0060 m

k= 9.00 × 10 ⁹ Nm²/C²

Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²

=2.8 × 10⁹ N

4 0
3 years ago
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The specific heat of a certain type of metal is 0.128 J/(g⋅∘C).0.128 J/(g⋅∘C). What is the final temperature if 305 J305 J of he
Makovka662 [10]

Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

T = 45.3°C

6 0
3 years ago
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A 4350 kg truck, driving 7.39 m/s, runs into the back of a stationary car. After the collision, the truck moves 4.55 m/s and the
DiKsa [7]

Answer:

Mass of car = 1098 kg

Explanation:

Here law of conservation of momentum is applied.

Let mass of car be m.

Initial momentum = Final momentum.

Initial momentum = 4350 x 7.39 + m x 0 = 32416.5 kgm/s

Final momentum = 4350 x 4.55 + m x 11.5 = 19792.5+11.5m

We have

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4 0
3 years ago
When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical
laila [671]

Answer:

I = 18 x 10⁻⁹ A = 18 nA

Explanation:

The current is defined as the flow of charge per unit time. Therefore,

I = q/t

where,

I = Average Current passing through nerve cell

q = Total flow of charges through nerve cell

t = time period of flow of charges

Here, in our case:

I = ?

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t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s

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<u>I = 18 x 10⁻⁹ A = 18 nA</u>

6 0
3 years ago
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