All categories

3 years ago

You are walking from your math class to your science class. You are carrying books

Physics

1 answer:

kolezko [41]3 years ago

8 0

Answer:

1800J

Explanation:

Given parameters:

Weight of the book = 20N

Total distance covered = 45m + 15m + 30m = 90m

Unknown:

Total work performed on the books = ?

Solution:

To solve this problem we must understand that work done is the force applied to move a body through a certain distance.

So;

Work done = Force x distance

Work done = 20 x 90 = 1800J

You might be interested in

A wave has a wavelength of 3.3 m and a speed of 5.6 m/s. What is the frequency of this wave

nikitadnepr [17]

Answer: λ (wavelength) = 3.3 m → ΔS = 3.3 m

v (speed) = 5.6 m/s → ΔV = 5.6 m/s

T (period) → ΔT = ?

f (frequency) = ?

If:

Now: (if, ΔT = T), Using the formula of the period of a wave, find the frequency:

Explanation: Your Welcome u.u

5 0

2 years ago

Which atmospheric layer prevents meteoroids from reaching earth’s surface

Natalija [7]

Stratosphere because the atmosphere get denser

5 0

3 years ago

A wire is made from a material having a temperature coefficient of resistivity of 0.0003125 (°C^-1). In an experiment, we mainta

beks73 [17]

Answer:

The power decreases by 36%

Explanation:

Given:

At 20° C

Power, P₀ = 300 W

Potential difference, V = 150 volts

Now, power is given as

P = V²/R

where, R is the resistance

on substituting the values, we get

300 = 150²/R₀

R₀ = 75 Ω

Now, the variation of resistance with temperature is given as

R = R₀[1 + α(T - T₀)]

where, α is the temperature coefficient of resistivity = 0.0003125 (°C⁻¹)

now, at

T₀ = 20° C

R₀ = 75 Ω

for

T = 1820° C

we have

R = R₀[1 + α(T - T₀)]

substituting the values

we get

R = 75×[1 + 0.0003125 × (1820 - 20)]

R = 117.18 Ω

Now using the formula for power

We have,

P = V²/R

P = 150²/117.18 = 192 W

Therefore, the percentage change will be

= $\frac{P-P_0}{P_0}\times 100$

on substituting the values , we get

= $\frac{192-300}{300}\times 100$

= -36%

here, negative sign depicts the decrease in power

3 0

3 years ago

Read 2 more answers

A charge q1= 3nC and a charge q2 = 4nC are located 2m apart. Where on the line passing through these charges is the total electr

Ipatiy [6.2K]

Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = $\frac{9\times 10^9\times3\times10^{-9}}{d^2}$

Field due to charge = 4 X 10⁻⁹ C

$E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$

These two fields will be equal and opposite to make net field zero

$\frac{9\times 10^9\times3\times10^{-9}}{d^2}$ = $[tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$ [/tex]

$\frac{3}{d^2} =\frac{4}{(2-d)^2}$

$\frac{2-d}{d} =\frac{2}{1.732}$

d = 0.928

5 0

3 years ago

A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a

trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden $d_1=1.1cm$

Speed of water in the hose is $v_1=0.95m/sec$

Number of holes n = 22

Diameter of each holes $d_2=15cm$

According to continuity equation $A_1v_1=A_2v_2$

$d_1^2\times v_1=22\times d_2^2v_2$

$1.1^2\times 0.95=22\times 0.15^2\times v_2$

$v_2=2.322m/sec$

So water leaves the sprinkler at a speed of 2.322 m/sec

6 0

3 years ago