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maria [59]
3 years ago
14

A child pulls a 3.00-kg sled across level ground at constant velocity with a light rope that makes an angle 30.0° above horizont

al. The tension in the rope is 5.00 N. Assuming the acceleration of gravity is 9.81 m/s2, what is the coefficient of friction between the sled and the ground?
Physics
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

μ = 0.161

Explanation:

Since the tension on the rope (F) is being applied at an angle (θ), it should be decomposed into a vertical and a horizontal component.

The vertical component of the tension on the rope (F*sin(θ)) is pulling the sled upwards, decreasing the normal force.

Therefore, the normal force can be calculated as:

N = m*g - F*sin(\theta)\\N= 3*9.8 - 5*sin(30)\\N= 26.9 N

The horizontal component of the tension on the rope (F*cos(θ)) is the force being applied to move the sled and that should equal the coefficient of friction multiplied by the normal force.

F*cos(\theta)=N*\mu\\5*cos(30)=26.9*\mu\\\mu=0.161

the coefficient of friction between the sled and the ground is 0.161

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For a given wave IF frequency doubles the wavelength
motikmotik

Answer:

C is halved

Explanation:

The frequency and the wavelength of a wave are related by the equation:

v=f\lambda

where

v is the speed of the wave

f is the frequency

\lambda is the wavelength

From the equation above, we see that for a given wave, if the wave is travelling in the same medium (and so, its speed is not changing), then the frequency and the wavelength are inversely proportional to each other.

Therefore, if the frequency doubles, the wavelength will halve in order to keep the speed constant:

\lambda = \frac{v}{f}\\\lambda' = \frac{v}{f'}=\frac{v}{2f}=\frac{1}{2}(\frac{v}{f})=\frac{\lambda}{2}

7 0
3 years ago
a lamp or bulb is marked 12 volt 240 Watts how many joules does it consume in one hour and what is the current that passes throu
valkas [14]

Answer:

1 watt is 1 Joule per second. 240 watt would then be 240 joules per second. How many seconds are there in an hour? Current formula is P= IV , P is 240(not the power in an hour because the current is calculated in seconds), and V is 12V. Then you will get I, the current.

5 0
3 years ago
What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

4 0
3 years ago
On January 22, 1943, the temperature in Spearfish, South Dakota, rose from -4.0∘F∘F to 45.0∘F∘F in just 2 minutes. What was the
Margaret [11]

Answer:

The change in temperature, \Delta T=9.45^{\circ} C

Explanation:

Given that,

The temperature in Spearfish, South Dakota, rose from -4^{\circ} F\ to\ 45^{\circ} F in just 2 minutes. We need to find the temperature change in Celsius degrees. Change in temperature is given by final temperature minus initial temperature such that,

\Delta T=T_f-T_i\\\\\Delta T=45-(-4)\\\\\Delta T=49^{\circ}F

The relation between degrees Celsius and degrees Fahrenheit is given by :

F=1.8C+32

Here, F = 49 degrees

49=1.8C+32\\\\\Delta T=9.45^{\circ} C

So, the change in temperature is 9.45 degree Celsius. Hence, this is the required solution.

8 0
3 years ago
Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia
Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

<u>Wn = 9.14 x 10¹⁷ N</u>

3 0
3 years ago
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