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3 years ago

Water that flows downhill along Earth's surface is called

1 answer:

8 0

<span>Water that flows downhill along the earth’s surface is called runoff. Runoff or also called as surface water runoff or overland flows is the flow of water when there’s excess water caused by ice melted, rain or storm. These occur because maybe the rain or the storm pours so hard and quick that the soil can’t absorb it fastly. If runoff water flows on the ground and reached a channel, it is called as nonpoint source. One example of nonpoint source are the canals full of leaves, the water can’t continue flowing because the leaves stops it.

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A space station, in the form of a spoked wheel 120 m in diameter, rotates to provide an artificial gravity of 3.00 m/s² for pers

lubasha [3.4K]

Answer:

2.12 rpm

Explanation:

$a_c= Centripetal\ acceleration=3.00\ m/s^2\\r=radius=\frac {120}{2}=60\ m\\a_c=\frac {v^2}{r}\\\Rightarrow v^2=a_c\times r\\\Rightarrow v^2=3\times 60=180\\\Rightarrow v=\sqrt{180}=13.41\ m/s\\$

$v=\omega r\\\Rightarrow \omega=\frac {v}{r}\\\Rightarrow \omega=\frac {13.41}{60}\\\Rightarrow \omega=0.2236\ rad/s\\1\ rad/s=9.55\ rpm\\\Rightarrow 0.2236\ rad/s=2.12\ rpm\\\therefore Wheel\ rotations=2.12\ rpm$

3 0

3 years ago

12. A boy of mass 50 kg running 5m/s jumps on to a 20kg trolley

atroni [7]

Answer:

4 m/s

Explanation:

Momentum is conserved.

m₁ v₁ + m₂ v₂ = (m₁ + m₂) v

(50)(5) + (20)(1.5) = (50 + 20) v

v = 4

The final velocity is 4 m/s.

7 0

3 years ago

A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow

shusha [124]

Answer:

150000000

$\dfrac{F_e}{F_g}=0.00000203873598369$

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

$1\ C=6.25\times 10^{18}\ electrons$

Number electrons is

$n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000$

The number of electrons added or removed was 150000000

Force is given by

$F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N$

The ratio is

$\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369$

The ratio is $\dfrac{F_e}{F_g}=0.00000203873598369$

Balancing the forces we get

$Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C$

The electric field required is 49050000 N/C

4 0

3 years ago

A car is initially moving at 20 m/s east and a little while later it is moving at 10 m/s north. Which of the following best desc

Nina [5.8K]

Answer:d

Explanation:

Given

First car is moving towards east with velocity 20 m/s

$\vec{v_1}=20\hat{i}$

then it turns towards north then velocity is

$\vec{v_2}=10\hat{j}$

suppose car takes t sec to change its path so average acceleration is given by

$a=\frac{v_2-v_1}{t}$

$a=\frac{1}{t}(10\hat{j}-20\hat{i})$

So average acceleration is towards North of west.

5 0

3 years ago

A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in

sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$ ---------------------- (1)

in which R₁ is the resistance at temperature T₁

$R_{T(2)$ is the resistance at temperature T₂

Given that V= IR

R = $\frac{V}{I}$

Therefore, the resistance at temperature 28°C is;

$R_{28}= \frac{120V}{1.36A}$

= 88.24Ω

$R_{T(2)$ = $\frac{120V}{1.23A}$

= 97.56Ω

From (1) above;

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

$\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)$

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

$\frac{0.1056}{4.5*10^{-4}}= T-28^0C$

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

$P= I^2_2R_{T^0C$

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0

3 years ago