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Dmitrij [34]
1 year ago
11

Junior slides across home plate during a baseball game. If he has a mass of 115 kg, and the coefficient of kinetic friction betw

een him and the ground is 0.35, what is the force of friction acting on him?
Physics
2 answers:
uysha [10]1 year ago
6 0

Answer:

like the other person said, it is 395

Explanation:

tatyana61 [14]1 year ago
4 0

Weight Force of Junior = m g = 115kg x 9.81 m/s^2 = 1128.15N then compute for the friction force


Friction Force= WF x (coefficient of kinetic friction) = 1128.15N x 0.35 =  394.8525N or 395N

 

But you can compute in a straightway:

Solution:

= 115 x 9.81 x 0.35

= 394.85

= 395 N

 

It will still give the same results.

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a student wants to push a box of books with the mass of 50 kg in 3 m horizontally towards the location of the shelves where the
irina1246 [14]

Answer:

The work done is 360 J.

Explanation:

Given that,

Mass = 50 kg

Distance =3 m

We need to calculate the work done

The work done is equal to the product of force and displacement.

Using formula of work done

W = F\cdot d

W = Fd\cos\theta

Where, F = force

D = distance

θ = Angle between force and displacement

Put the value into the formula

W=120\times3\cos0^{\circ}

W=360\ J

Hence, The work done is 360 J.

8 0
2 years ago
A magnetic dipole with a dipole moment of magnitude 0.0243 J/T is released from rest in a uniform magnetic field of magnitude 57
ololo11 [35]

Answer:

47.76°

Explanation:

Magnitude of dipole moment = 0.0243J/T

Magnetic Field = 57.5mT

kinetic energy = 0.458mJ

∇U = -∇K

Uf - Ui = -0.458mJ

Ui - Uf = 0.458mJ

(-μBcosθi) - (-μBcosθf) = 0.458mJ

rearranging the equation,

(μBcosθf) - (μBcosθi) = 0.458mJ

μB * (cosθf - cosθi) = 0.458mJ

θf is at 0° because the dipole moment is aligned with the magnetic field.

μB * (cos 0 - cos θi) = 0.458mJ

but cos 0 = 1

(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³

1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³

1 - cos θi = 0.3278

collect like terms

cosθi = 0.6722

θ = cos⁻ 0.6722

θ = 47.76°

7 0
2 years ago
Read 2 more answers
Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp
UkoKoshka [18]

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

6 0
2 years ago
In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
scoray [572]

Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal  force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

F = 2511.12 N

net vertical force:

F sin 30° - T sin 19°- W = 0

W = F sin 30° - T sin 19°

W =  2511.12 sin 30° - 2300 sin 19°

W = 506.75 N

8 0
2 years ago
A spaceship flies from Earth to a distant star at a constant speed. Upon arrival, a clock on board the spaceship shows a total e
m_a_m_a [10]

Answer:

35 288 mile/sec

Explanation:

This is a problem of special relativity. The clocks start when the spaceship passes Earth with a velocity v, relative to the earth. So, out and back from the earth it will take:

10 years = \frac{2d}{v}

If we use the Lorentz factor, then, as observed by the crew of the ship, the arrival time will be:

0.8 = \sqrt{1-\frac{v^{2} }{c^{2} } }

Then the amount of time wil expressed as a reciprocal of the Lorentz factor. Thus:

0.8 = \sqrt{1 - \frac{v^{2} }{c^{2} } }

0.64 = 1-\frac{v^{2} }{186282^{2} }

solving for v, gives = 35 288 miles/s

4 0
2 years ago
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