Doubling the voltage in a circuit doubles the current if what is held constant?

A simple pendulum bhas a period of 3.45s when the length of the pendulum is shortened by 1m the period is 2.81s caculate the ori

Bond [772]

Chloe?????????????.......

7 0

3 years ago

zanele has a mass of 40kg and is sitting inside a 20kg cart .zanele's friends pull a cart with a force of 500N at an angle of 20

Dmitriy789 [7]

The net horizontal force acting on the cart is 169.85 N.

The change in the carts momentum is 5,000 Ns.

The net horizontal force on Zanele is same as the net horizontal force on the cart.

If angle between the 500N force and the horizontal is decreased, the final velocity of the cart increases.

The given parameters:

Mass of zanele, m1 = 40 kg
Mass of the cart, m2 = 20 kg
Applied force, F = 500 N, at angle 20 degrees
Frictional force on the cart, Ff = 300 N
Time, t = 10 s

<h3>Net horizontal force on the cart;</h3>

The net horizontal force acting on the cart is calculated as follows;

$F_{net} = F - F_f\\\\F_{net} = 500 \times cos(20) - 300\\\\F_{net} = 169.85 \ N$

<h3>Change in momentum;</h3>

The change in the carts momentum at the given time of applied force is calculated as follows;

$\Delta mv = Ft \\\\\Delta mv = 500 \times 10= 5,000 \ Ns$

The net horizontal force on Zanele is same as the net horizontal force on the cart.

<h3>The final velocity of the cart;</h3>

When the angle decreases the cart's final velocity would be affected as follows;

$F_{net} = F - F_f\\\\F_{net} = \frac{mv}{t} \\\\\frac{mv}{t} = F - F_f\\\\ \frac{mv}{t} = Fcos(\theta) - F_f\\\\let \ \theta = 0^0\\\\\frac{mv}{t} =500 \times cos(0) - 300\\\\\frac{mv}{t} = 200\\\\v = \frac{200 t}{m} \\\\when \ \theta = 20^0\\\\\frac{mv}{t} =500 \times cos(20) - 300\\\\\frac{mv}{t} = 169.85\\\\v = \frac{169.85t}{m}$

Thus, if angle between the 500N force and the horizontal is decreased, the final velocity of the cart increases.

Learn more about net horizontal force here: brainly.com/question/21684583

5 0

3 years ago

A 14.0 kg cart is coasting down a street at a steady speed of 7.00 m/s. what is the kinetic energy of the cart?

Orlov [11]

I had this question before can you show me the answer choices

3 0

4 years ago

What are the 3 laws of Newton

daser333 [38]

Answer:

What are Newton's 1st 2nd and 3rd laws of motion?

They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. More precisely, the first law defines the force qualitatively, the second law offers a quantitative measure of the force, and the third asserts that a single isolated force doesn't exist.

Explanation:

They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. More precisely, the first law defines the force qualitatively, the second law offers a quantitative measure of the force, and the third asserts that a single isolated force doesn't exist.

5 0

4 years ago

A 1.50 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

Tema [17]

Answer:

α = 1.114 × 10⁻³ (°C)⁻¹

Explanation:

Given that:

Length of rod (L) = 1.5 m,

Diameter (d) = 0.55 cm,

Area (A) = $\pi r^2$

Radius (r) = d / 2 = 0.275 cm,

Voltage across the rod (V) = 15.0 V.

At initial temperature (T₀) = 20°C, the current (I₀) = 18.8 A while at a temperature (T) = 92⁰C, the current (I) = 17.4 A

a) The resistance of the rod (R) is given as:

$R=\frac{Voltage(V)}{I_0} \\R=\frac{15}{18.8}=0.798\Omega$

Therefore the resistivity and for the material of the rod at 20 °C (ρ) is:

$\rho=\frac{RA}{L}=\frac{0.798*\pi *0.275^2}{1.5}=0.126\Omega m$

b) The temperature coefficient of resistivity at 20°C for the material of the rod (α) can be gotten from the equation:

$R_T=R_0[1-\alpha (T-T_0)]\\but,R_T=\frac{V}{I}=\frac{15}{17.4}=0.862\\$

Rearranging to make α the subject of formula:

$\frac{R_T}{R_0} =1+\alpha (T-T_0)\\\alpha (T-T_0)=\frac{R_T}{R_0}-1\\\alpha =\frac{\frac{R_T}{R_0}-1}{(T-T_0)} \\Substituting:\\\alpha =\frac{\frac{0.862}{0.798}-1 }{92-20} \\\alpha =\frac{0.0802}{72} =1.114*10^-3(^0C)^{-1$

8 0

3 years ago