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kap26 [50]
3 years ago
5

A lawn roller is rolled across a lawn by a force of 107 N along the direction of the handle, which is 13.5 ◦ above the horizonta

l. If the lawn roller develops 69.4 W of power for 33 s, what distance is the roller pushed? Answer in units of m.
Physics
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer:

22.02 m

Explanation:

given,

Force, F = 107 N

angle made with horizontal = 13.5◦

Power develop by the lawn roller = 69.4 W

time = 33 s

distance = ?

Force along horizontal= F cos θ

          = 107 cos 13.5°= 104 N

Power = \dfrac{work\ done}{time}

69.4 = \dfrac{W}{33}

W = 2290.2 J

Work done= Force x displacement

displacement= \dfrac{2290.2}{104}

                      = 22.02 m

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Ratio of resistances of two bulbs is 2:3. If they are connected in series to a supply, then the ratio of voltages across them is
V125BC [204]

Answer:

Explanation:

Given that,

Two resistor has resistance in the ratio 2:3

Then,

R1 : R2 = 2:3

R1 / R2 =⅔

3 •R1 = 2• R2

Let R2 = R

Then,

R1 = ⅔R2 = 2/3 R

So, if the resistor are connected in series

Let know the current that will flow in the circuit

Series connection will have a equivalent resistance of

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Req = R + ⅔ R = 5/3 R

Req = 5R / 3

Let a voltage V be connect across then, the current that flows can be calculated using ohms law

V = iR

I = V/Req

I = V / (5R /3)

I = 3V / 5R

This the current that flows in the two resistors since the same current flows in series connection

Now, using ohms law again to calculated voltage in each resistor

V= iR

For R1 = ⅔R

V1 =i•R1

V1 = 3V / 5R × 2R / 3

V1 = 3V × 2R / 5R × 3

V1 = 2V / 5

For R2 = R

V2 = i•R2

V2 = 3V / 5R × R

V2 = 3V × R / 5R

V2 = 3V / 5

Then,

Ratio of voltage 1 to voltage 2

V1 : V2 = V1 / V2 = 2V / 5 ÷ 3V / 5

V1 : V2 = 2V / 5 × 5 / 3V.

V1 : V2 =2 / 3

V1:V2 = 2:3

The ratio of their voltages is also 2:3

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3 years ago
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The electric field intensity generated by a single point charge is given by

E(r)=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge


In this problem, the charge is q=25 \mu C=25 \cdot 10^{-6} C and we are asked to calculate the field at distance r=9.0 m, so the electric field is

E(9.0 m)=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{25 \cdot 10^{-6} C}{(9 m)^2}=2775 V/m

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While practicing the trumpet you notice that every time you play a particular note a window in the room rattles. How can you exp
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3 years ago
One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached t
Jobisdone [24]

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2\frac{m}{s^{2} }

Explanation:

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To determine, the acceleration we solve the following equation for a:

F=ma\\a=\frac{F}{m} =\frac{51.0N}{5.00kg}\\a=10.2\frac{m}{s^{2} }

8 0
4 years ago
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