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dezoksy [38]
3 years ago
10

25 Points PLEASE I need Help

Physics
1 answer:
olga55 [171]3 years ago
6 0

Answer:nah u took my points I take urs

Explanation:

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A convex mirror has a focal length of -10.8 cm. An object is placed 32.7 cm from the mirror's surface. Determine the image dista
KonstantinChe [14]

Answer:

-353.16

Explanation:

4 0
3 years ago
Plz help This is for Rock formations...only 5 sentences if not more
dsp73

Answer:

The coastal zone is not a stable and constant environment, but a dynamic place that can change rapidly in response to natural processes such as seasonal weather patterns. Waves, winds, currents, tides and storms are the major forces on the coast.

Explanation:

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3 years ago
A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h
timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

F_n = \frac{mv^2}{r}

N+mgcos\theta = \frac{mv^2}{r}

Here,

Normal reaction of the ring is N and velocity of the ring is v

N+mgcos\theta = \frac{mv^2}{r}

N+Wcos\theta = \frac{W}{g} (\frac{v^2}{r})

N+2cos30\° = \frac{2}{32.2}*\frac{10^2}{2}

N = 1.374lb

PART B) Acceleration

F_t = ma_t

-mgsin\theta = ma_t

-W sin\theta = \frac{W}{g} a_t

-2Sin30\° = (\frac{2}{32.2})a_t

a_T = -16.10ft/s^2

Negative symbol indicates deceleration.

<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>

8 0
3 years ago
A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.
ValentinkaMS [17]

A) 2.03 m/s^2

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

mg sin \theta - \mu_k R = ma (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

\theta=22.5^{\circ}

\mu_k = 0.19 is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

R-mg cos \theta =0 (2)

From (2) we find

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2

B) 5.94 m/s

We can solve this part by using the suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

a=2.03 m/s^2

s = 8.70 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s

6 0
3 years ago
A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and th
gizmo_the_mogwai [7]

Answer:

λ = 5.2 x 10⁻⁷ m = 520 nm

Explanation:

From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:

Δx = λL/d

where,

Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m

L = Distance between slits and screen = 3.1 m

d = Separation between slits = 0.0005 m

λ = wavelength of light = ?

Therefore,

0.00322 m = λ(3.1 m)/(0.0005 m)

λ = (0.00322 m)(0.0005 m)/(3.1 m)

<u>λ = 5.2 x 10⁻⁷ m = 520 nm</u>

5 0
3 years ago
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