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elena55 [62]
3 years ago
14

A ______ is something that can be placed between the sun and the subject to diffuse the light. *

Physics
2 answers:
Gekata [30.6K]3 years ago
6 0

Answer:

scrim

Explanation:

A scrim is something that can be placed between the sun and the subject to diffuse the light.

An instance of a diffuser is a softbox that is put on its front side around a strobe containing diffusion content. The sun is a form of hard light that is often diffused through a scrim. The light rays are dispersed by putting a scrim between the sun and the object, and the harsh sun's rays is gentler.

nataly862011 [7]3 years ago
6 0

Answer:

A: Flag

Explanation:

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A dog has a weight of 700N find the mass of the dog
Stolb23 [73]
The specific answer for that will be 71.38 kg
8 0
3 years ago
Two blocks are connected by a light weight, flexible cord that passes over a frictionless pulley.Ifm1=2 kg and m2 = 3 kg, and bl
Vera_Pavlovna [14]

Answer:

t = 1.41 sec.

Explanation:

If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.

First, we need to find the value of  acceleration, which is the same for both blocks.

If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:

F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)

⇒ a = (\frac{(m₂-m₁)}({m₁+m₂} * g = g/5 m/s²

Once we got the value of a, we can use for instance this kinematic equation, and solve for t:

Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.

6 0
3 years ago
A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc
Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0
3 years ago
Charges q3 and q5 are now replaced by two charges, q2 and q6, having equal magnitude and sign (-3.4 μC). Charge q2 is located at
ivanzaharov [21]
I believe the answer would be zero because the q1 and q2 are equal on opposite sides and it

hope this helps
3 0
3 years ago
What is the potential energy of a 1000 kg-ball that is on the ground?
Kazeer [188]

Answer:

0J

Explanation:

PE=mgh

PE=  1000kgx9.8m/s^2x0m

PE = 0J

6 0
2 years ago
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