All of the following are types of telescope except

the goat weighs 900 N and is 1 meter from the fulcrum. The strongman pulls down on the lever 3 meters from the fulcrum. What is

garik1379 [7]

Answer: The smallest effort = 300N

Explanation:

Using one of the condition for the attainment of equilibrium:

Clockwise moment = anticlockwise moments

900 × 1 = 3 × M

Where M = the weight of the strong man

3M = 900

M = 900/3 = 300N

Therefore, 300N is the smallest effort that the strongman can use to lift the goat

6 0

3 years ago

If you shine a light of frequency 375hz on a double slit setup , and you measure the slit separation to be 950 nm and the screen

amm1812

Answer:

y = 33.93 10⁵ m

Explanation:

This is an interference exercise, for the contributory interference is described by the expression

d sin θ = m λ

let's use trigonometry for the angle

tan θ = y / L

how the angles are small

tan θ = sin θ / cos tea = sin θ

we substitute

sin θ = y / L

d y / L = m λ

y = m λ L / d

the light fulfills the relation of the waves

c = λ f

λ = c / f

λ = 3 10⁸ /375

λ = 8 10⁵ m

first order m = 1

let's calculate

y = 1 8 10⁵ 4030 10-9 / 950 10-9

y = 33.93 10⁵ m

6 0

3 years ago

Margaret wants to go for a swim, and decides to jump in using the diving board that measures 3-m long.

otez555 [7]

Answer:

The horizontal component of her velocity is approximately 1.389 m/s

The vertical component of her velocity is approximately 7.878 m/s

Explanation:

The given question parameters are;

The initial velocity with which Margaret leaps, v = 8.0 m/s

The angle to the horizontal with which she jumps, θ = 80° to the horizontal

The horizontal component of her velocity, vₓ = v × cos(θ)

∴ vₓ = 8.0 × cos(80°) ≈ 1.389

The horizontal component of her velocity, vₓ ≈ 1.389 m/s

The vertical component of her velocity, $v_y$ = v × sin(θ)

∴ $v_y$ = 8.0 × sin(80°) ≈ 7.878

The vertical component of her velocity, $v_y$ ≈ 7.878 m/s.

6 0

3 years ago

Problem 3: Thermal expansionThe steel rod has the length 2 m and cross-section area 200 cm2at the room temperature 20◦C. Weapply

zalisa [80]

Answer:

a) 2.00024 m

b) 0.036%

c) 436.67°C

Explanation:

Given

Initial length = L₀ = 2 m

Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³

Initial Temperature = T₀ = 20°C

Coefficient of linear expansivity = α = (2 × 10⁻⁶) (°C)⁻¹

a) New length of the rod after heating to 80°C

Linear expansion is given as

ΔL = L₀ × α ×ΔT

ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm

New length = old length + expansion = 2 + 0.00024 = 2.00024 m

b) The percentage of the volume change of the rod.

Volume expansion is given by

ΔV = V₀ × (3α) × ΔT

Volume expansivity ≈ 3 × (linear expansivity)

ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³

Percentage change in volume = 100% × (ΔV/V₀) = 100% × (0.0000144/0.04) = 0.036%

c) The maximal temperature we can allow if the volume should not increase by more than half percent.

For a half percent increase in volume, the corresponding change in volume needs to be first calculated.

Percentage change in volume = 100% × (ΔV/V₀)

0.5 = 100% × (ΔV/0.04)

(ΔV/0.04) = 0.005

ΔV = 0.0002 m³

Then we now investigate the corresponding temperature that causes this.

ΔV = V₀ × (3α) × ΔT

0.0002 = 0.04 × (3×2×10⁻⁶) × ΔT

ΔT = (0.0002)/(0.04 × 3 × 2 × 10⁻⁶) = 416.67°C

Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C

4 0

2 years ago

A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$