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Anton [14]
3 years ago
9

James accelerates his skate board uniformly along a straight road from rest to 10 m/s in 4 seconds. What is James Acceleration?

Physics
1 answer:
lawyer [7]3 years ago
4 0

Given:

u(initial velocity)=0

v(final velocity)= 10 m/s

t= 4 sec

Now we know that

v= u + at

Where v is the final velocity

u is the initial velocity

a is the acceleration measured in m/s^2

t is the time measured in sec

10=0+ax4

a=10/4

a=2.5 m/s^2

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A 12 kg<br> mass is lifted to a height of 2 m. What is its potential energy<br> at this position?
Romashka-Z-Leto [24]

Answer:

Explanation:

Potential energy is the energy stored within an object, due to the object's position, arrangement or state

4 0
3 years ago
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If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds
BabaBlast [244]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

According to above question ~

  • Current (I) = 4 Amperes

  • Time (t) = 3 seconds

  • Charge (q) = ?

Let's find the charge (q) by using formula ~

  • I =  \dfrac{q}{t}

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  • q = 4 \times 3

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Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds

7 0
3 years ago
A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the
blagie [28]

Answer: q2 = -0.05286

Explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

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Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

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3 years ago
How many sig figs are in 3252.6
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The number 3252.6 has 5 significant figures
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A raft with the area A , thickness= h and the mass 600 kg, Floats in still water with 7 cm
elena55 [62]

<span>In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, </span>where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.

<span>ρwAd1g = Mg</span>

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<span>d2∕d1 = (M + m)/g</span>

m = [(d2<span>∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg</span>

This means that Bubba’s mass is 120 kg.

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