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2 years ago

What is the absolute value of the horizontal force that each athlete exerts against the ground?

Physics

1 answer:

alexandr402 [8]2 years ago

6 0

Refer to the diagram shown below.

When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.

At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.

Answer:
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.

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Answer:

KE= 1/2mv²

Explanation:

The kinetic energy of a body is the energy possessed by virtue of the body in motion

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The expression for the kinetic energy of a body is given as

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2 years ago

A stationary particle of charge q = 2.1 × 10-8 c is placed in a laser beam (an electromagnetic wave) whose intensity is 2.9 × 10

alisha [4.7K]

(a) The intensity of the electromagnetic wave is related to the amplitude of the electric field by
$I= \frac{1}{2} c \epsilon_0 E^2$
where
I is the intensity
c is the speed of light
$\epsilon_0$ is the electric permittivity
E is the amplitude of the electric field

By substituting the numbers of the problem and re-arranging the equation, we can find E:
$E= \frac{2 I}{c \epsilon_0} = \frac{2 ( 2.9 \cdot 10^3 Wm^{-2})}{(3 \cdot 10^8 m/s)(8.85 \cdot 10^{-12} Fm^{-1})} =2.2 \cdot 10^6 N/C$

Now that we have the intensity of the electric field, we can calculate the electric force on the charge:
$F=qE=(2.1 \cdot 10^{-8} C)(2.2 \cdot 10^6 N/C)=0.046 N$

(b) We can calculate the amplitude of the magnetic field starting from the amplitude of the electric field:
$B= \frac{E}{c}= \frac{2.2 \cdot 10^6 N/C}{3 \cdot 10^8 m/s}=7.3 \cdot 10^{-3} T$

The magnetic force is given by
$F=qvB \sin \theta$
where v is the particle's speed, B the magnetic field intensity and $\theta$ the angle between B and v.
In this case the charge is stationary, so v=0, and so the magnetic force is zero: F=0.

(c) The electric force has not changed compared to point (a), because it does not depend on the speed of the particle, so we have again F=0.046 N.

(d) This time, the particle is moving with speed $v=3.7 \cdot 10^4 m/s$ , in a direction perpendicular to the magnetic field (so, the angle $\theta$ is $90^{\circ}$ ), and so by using the intensity of the magnetic field we found in point (b), we can calculate the magnetic force on the particle:
$F=qvB \sin \theta = (2.1 \cdot 10^{-8}C)(3.7 \cdot 10^4 m/s)(7.3 \cdot 10^{-3} T)(\sin 90^{\circ} )=$
$=5.7 \cdot 10^{-6} N$

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2 years ago

What is the magnitude of the force a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away?

sweet [91]

The magnitude of the force a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away is 1920 Newtons. The formula used to solve this problem is:

F = kq1q2/r^2

where:
F = Electric force, Newtons
k = Coulomb's constant, 9x10^9 Nm^2/C^2
q1 = point charge 1, C
q2 = point charge 2, C
r = distance between charges, meters

Using direct substitution, the force F is determined to be 1920 Newtons.

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2 years ago

The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan

PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire. If the current triples while the distance doubles, the strength of the magnetic field increases by one and half (1.5) times.

Explanation:

Magnetic field around a long current carrying wire is given by

$B=\frac{\mu _o I}{2\pi r}$

where B= magnetic field

$\mu _o=$ permeability of free space

I= current in the long wire and

r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire.

Now if I'=3I and r'=2r then magnetic field B' is given by

$B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B$

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by one and half (1.5) times.

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2 years ago

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A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0

Serjik [45]

Answer:

The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.

Explanation:

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$F_{lift}=F_{up} - F_{dw}=0.5*p*V^2$

Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.

When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:

$F_{lift}=0.5*p*V^2=W$