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Alex17521 [72]
3 years ago
12

An initially stationary 2.7 kg object accelerates horizontally and uniformly to a speed of 13 m/s in 4.0 s. (a) In that 4.0 s in

terval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?
Physics
1 answer:
Ronch [10]3 years ago
8 0

Explanation:

A.

Given:

V = 13 m/s

t = 4 s

Constant acceleration, a= (V-Vi)/t

= 13/4

= 3.25 m/s^2

F = mass * acceleration

= 2.7 * 3.25

= 8.775 N.

Using equations of motion,

distance,S = (13 * 4) - (1/2)(3.25)(4^2)

= 26 m

Workdone, W = force * distance

= 8.775 * 26

= 228.15 J

B.

Instantaneous power, P = Force *Velocity

= 8.775 * 13

= 114. 075 W

C.

t = 2 s,

Constant acceleration, a= (V-Vi)/t

= 13/2

= 6.5 m/s^2

Force = mass * acceleration

= 2.7 * 6.5

= 17.55 N

Instantaneous power, P = Force *Velocity

= 17.55 * 13

= 228.15 W.

= 114. 075 W.

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23.
Volgvan

Answer:

1 m = 100 cm....so 2.5 m = (2.5 * 100) = 250 cm

a = 1st shelf

b = 2nd

c = 3rd

d = 4th

a + b + c + d = 250

b = 2a + 18

c = a - 12

d = a + 4

a + (2a + 18) + (a - 12) + (a + 4) = 250

5a + 10 = 250

5a = 250 - 10

5a = 240

a = 240/5

a = 48 cm <== 1st shelf

b = 2a + 18 = 2(48) + 18 = 114 cm <== 2nd shelf

c = a - 12 = 48 - 12 = 36 cm <== 3rd shelf

d = a + 4 = 48 + 4 = 52 cm <== 4th shelf

so 2nd shelf is 114 cm

4 0
2 years ago
Need help with this physics question!
Fynjy0 [20]

Answer:

omg i need help with the same answer lol

Explanation:

i wish i can help but i need help on this hehe

6 0
3 years ago
If the density of an object is 5.2 g/cm3, and volume is 3.7 cm3, what is its mass
netineya [11]
Here's the equation you use: Density = mass/volume

1) 5.2g/cm^3 = m/3.7cm^3

2) m = 5.2g/cm^3 x 3.7cm^3

3) m = 19.24g

You can check the answer by plugging it in

19.24g/3.7cm^3
= 5.2g/cm^3
6 0
3 years ago
Read 2 more answers
Three point charges are arranged along the x-axis. Charge q1 = +3.00 uC is at the origin, and charge q2= -5.00 uC is at x= 0.200
artcher [175]
We have all the charges for q1, q2, and q3. 
Since k = 8.988x10^2, and N=m^2/c^2

F(1) = F (2on1) + F (3on1)

F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 |   / (.2m)^2
F(2on1) = 3.37 N

Since F1 is 7N,

F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)

Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N

F(3on1) = k |q1 q3| / r(the distance between the two)^2 
r^2 x F(3on1) = k |q1 q3| 
r = sqrt of k |q1 q3| / F(3on1) 
= .144 m (distance between q1 and q3)
0 - .144m 

So it's located in -.144m

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
6 0
3 years ago
Read 2 more answers
How much heat is required to heat 2 kg of water from 25°C to 40°C?
Dominik [7]

Answer:

126000 J

Explanation:

Applying,

Q = cm(t₂-t₁).................. Equation 1

Where Q = Amount of heat, c = specifc heat capacity of water, m = mass of water, t₁ = Initial temperature, t₂ = Final temperature.

From the question,

Given: m = 2 kg, t₁ = 25°C, t₂ = 40°C

Constant: c = 4200 J/kg.°C

Substitute these value into equation 1

Q = 2×4200(40-25)

Q = 2×4200×15

Q = 126000 J

5 0
3 years ago
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