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3 years ago

An object with a height of 1.0 cm is placed 1.4 cm from a convex lens that has a

Physics

1 answer:

Gennadij [26K]3 years ago

5 0

Answer:

A concave mirror has a focal length of. 10.0 cm. What is its radius of curvature? ... 20.0 cm. 62. An object located 18 cm from a convex mirror produces a virtual image 9 ... cm. 75 cm. 66. Find the image position and height for the object shown in ... 1 block 1.0 cm. Vertical scale: 2 blocks 1.0 cm. F. I1 hi. 1.0 cm di. 2.7 cm. O1.

Explanation:

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Why are light intensity, carbon dioxide concentration and temperature liimiting factors

Harman [31]

Answer:

As carbon dioxide concentrations increase, so too does the rate of photosynthesis until a certain point where the graph levels off. At lower carbon dioxide concentrations carbon dioxide is the limiting factor because an increase in carbon dioxide causes an increase in photosynthesis.

Explanation:

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3 years ago

Enter an expression, in terms of defined quantities and g, for the force that the scale under left pillar shows

ella [17]

The expression, in terms of defined quantities and g is therefore Fu =((mg/2) +2 mp) g

<h3>What is a Scale?</h3>

This can be defined as a balance or any of various other instruments or devices for weighing.

The expression in terms of defined quantities and g, for the force that the scale under left pillar shows that Fu =((mg/2) +2 mp) g .

Read more about Force here brainly.com/question/4515354

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2 years ago

A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.

SVETLANKA909090 [29]

A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of $x_0 = 5.0 m$

- at t = 0, the initial velocity of the motorcyclist is $v_0 = 15.0 m$ east

- The acceleration of the motorcyclist is constant and it is $a=4.0 m/s^2$

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

$x(t)=x_0 + v_0t + \frac{1}{2}at^2$

where t is the time.

Substituting t = 2.0 s, we find the position:

$x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m$

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

$v(t)=x'(t)=v_0 + at$

where:

$v_0=15.0 m/s$ is the initial velocity

$a=4.0 m/s^2$ is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

$t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s$

And therefore, the position at t = 2.5 s is:

$x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m$

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3 0

3 years ago

The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s

alex41 [277]

Given:

Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.

We need to determine the maximum shear stress developed in the beam:

τ = F/A

Assuming the area of the beam is 100 m^2 with a length of 10 m.

τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />

8 0

3 years ago

Pls send the answer..pls...

Sophie [7]

Answer:

The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

Explanation:

Given;

initial temperature of the liquid, t₁ = 76.3 +/- 0.4⁰C

final temperature of the liquid, t₂ = 67.7 +/- 0.3⁰C

The change in temperature of the liquid is calculated as;

Δt = t₂ - t₁

Δt = (67.7 - 76.3) +/- (0.3 - 0.4)

Δt = (-8.6) +/- (-0.1)

Δt = 8.6 +/- 0.1 ⁰C

Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

4 0

3 years ago