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3 years ago

Alan slides across home plate during a baseball game. If he has a mass of 61

Physics

2 answers:

andrew11 [14]3 years ago

7 0

Answer:

D. 371 N

Explanation:

Friction force equals normal force times the coefficient of friction:

F = Nμ

Since the ground is level, normal force equals weight:

N = mg

Therefore:

F = mgμ

F = (61 kg) (9.8 m/s²) (0.62)

F = 371 N

viktelen [127]3 years ago

3 0

Answer:

371 N

Explanation:

A P E X Verified

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Astronomers know that the distance between the Earth and the Sun averages 1.50 x108 km. How can astronomers use the observed ste

rodikova [14]

Answer:

The distance of stars and the earth can be averagely measured by using the knowledge of geometry to estimate the stellar parallax angle(p).

From the equation below, the stars distances can be calculated.

D = 1/p

Distance = 1/(parallax angle)

Stellar parallax can be used to determine the distance of stars from an observer, on the surface of the earth due to the motion of the observer. It is the relative or apparent angular displacement of the star, due to the displacement of the observer.

Explanation:

Parallax is the observed apparent change in the position of an object resulting from a change in the position of the observer. Specifically, in the case of astronomy it refers to the apparent displacement of a nearby star as seen from an observer on Earth.

The parallax of an object can be used to approximate the distance to an object using the formula:

D = 1/p

Where p is the parallax angle observed using geometry and D is the actual distance measured in parsecs. A parsec is defined as the distance at which an object has a parallax of 1 arcsecond. This distance is approximately 3.26 light years

3 0

3 years ago

The sensitivity of a measuring instrument is the value of the smallest quantity that can be read or estimated with it. What is t

nirvana33 [79]

Answer:

The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.

6 0

3 years ago

Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,

yarga [219]

Answer:

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

= (9×10^9)×(q)/(24.0)

= 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

R^2 = 9×r^2

E2 = k×q/(9×r^2)

= k×q/(9×375000000q)

= k/(9×375000000)

= (9×10^9)/(9×375000000)

= 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

5 0

3 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago

Hi im a little stuck on this question that came from my textbook in class it got a little confusing for me because i really dont

koban [17]

Consult the attached free body diagram.

By Newton's second law, the net force on the crate acting parallel to the surface is

∑ F[para] = (370 N) cos(-20°) - f = 0

(this is the x-component of the resultant force)

where

• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force

• f = magnitude of kinetic friction

The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.

Solve for f :

f = (370 N) cos(-20°) ≈ 347.686 N

The net force acting perpendicular to the surface is

∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0

(this is the y-component of the resultant force)

where

• n = magnitude of normal force

• 1480 N = weight of the crate

• (370 N) sin(-20°) = magnitude of the vertical component of push

The crate doesn't move up or down, so it's also in equilibrium in this direction.

Solve for n :

n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N

Then the coefficient of kinetic friction is µ such that

f = µn ⇒ µ = f/n ≈ 0.216

7 0

2 years ago