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2 years ago

PLEASE HELPPPPPP ME PLS PLS

Physics

1 answer:

Marta_Voda [28]2 years ago

6 0

Power = work / time --> time = work / power = 3600 J / 275 watts = 13.1 seconds.

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A solenoid 0.425 m long has 950 turns of wire. What is the magnetic field in the center of the solenoid when it carries a curren

lesya692 [45]

Answer:

The magnetic field in the center of the solenoid is $7.8\times10^{-3}\ T$ .

Explanation:

Given that,

Length of solenoid = 0.425 m

Number of turns N = 950

Current I = 2.75 A

The magnetic field in the center of the solenoid is the product of the current , number of turns per unit length and permeability.

In mathematical form,

$B = \mu_{0}nI$

Where, $n = \dfrac{N}{l}$

N = number of turns

L = length

I = current

Now, The magnetic field

$B = \dfrac{\mu_{0}NI}{l}$

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times950\times2.75}{0.425}$

$B=\dfrac{4\times3.14\times10^{-7}\times950\times2.75}{0.425}$

$B=7.8\times10^{-3}\ T$

Hence, The magnetic field in the center of the solenoid is $7.8\times10^{-3}\ T$ .

4 0

3 years ago

What is relationship between latent heat and water ?

Wittaler [7]

Latent heat, energy absorbed or released by a substance during a change in its physical state (phase) that occurs without changing its temperature.

5 0

3 years ago

What electrical force dies a Uranium nucleus exert on one of its inner electrons, located at a distance of 175 picometers (= 1.7

AlekseyPX

Answer:

correct option is d) 7.0 x 10^-7 N

Explanation:

given data

distance = 175 picometers = 1.75 × $10^{-10}$ m

to find out

electrical force

solution

we know atomic no of uranium is 92

and charge on electron is = 1.6 × $10^{-19}$ C

and electrical force is express as

electrical force = $\frac{1}{4 \pi \epsilon _o} \frac{q1q2}{r^2}$ .............1

put here value we get

electrical force = $9*10^9 \frac{92*(1.6*10^{-19})^2}{(1.75*10^{-10})^2}$

electrical force = 6.921 × $10^{-7}$ N

so correct option is d) 7.0 x 10^-7 N

5 0

3 years ago

Read 2 more answers

A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio

Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

$F$ - External force exerted on the sled, measured in newtons.

$f$ - Friction force, measured in newtons.

$N$ - Normal force from the ground on the mass, measured in newtons.

$W$ - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

$W = m\cdot g$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 50\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the weight of the sled is:

$W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$W = 490.35\,N$

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

$F_{min,s} = \mu_{s}\cdot W$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

$W$ - Weight of the sled, measured in newtons.

If we know that $\mu_{s} = 0.3$ and $W = 490.35\,N$ , then the force needed to start the sled moving is:

$F_{min,s} = 0.3\cdot (490.35\,N)$

$F_{min,s} = 147.105\,N$

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

$F_{min,k} = \mu_{k}\cdot W$ (3)

Where $\mu_{k}$ is the kinetic coefficient of friction, dimensionless.

If we know that $\mu_{k} = 0.1$ and $W = 490.35\,N$ , then the force needed to keep the sled moving at a constant velocity is:

$F_{min,k} = 0.1\cdot (490.35\,N)$

$F_{min,k} = 49.035\,N$

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

8 0

2 years ago

g (12 points) The time between incoming phone calls at a call center is a random variable with exponential density p(x) = 1 r e

rusak2 [61]

Answer:

$(1)p(x)\geq 0\\(2)\int_{0}^{\infty} p(x) dx=0$

Explanation:

A function f(x) is a Probability Density Function if it satisfies the following conditions:

$(1)f(x)\geq 0\\(2)\int_{0}^{\infty} f(x) dx=0$

Given the function:

$p(x)=\dfrac{1}{r}e^{-x/r} \: on\: [0,\infty), where\:r=\dfrac{20}{ln(2)}$

(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in $[0,\infty).$

(2)

$\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1$

The function p(x) satisfies the conditions for a probability density function.

6 0

3 years ago

PLEASE HELPPPPPP ME PLS PLS​

PLEASE HELPPPPPP ME PLS PLS