1.si un automovil de 3000 kg se desplaza a 40 m/s su energía cinética es igual a

consider the mirror from the last question. an object 4cm tall stands 10cm in front of a converging mirror of focal length 5cm.

aleksandr82 [10.1K]

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6 0

3 years ago

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When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, th

Fofino [41]

Answer:

Option E is correct 310N

Explanation:

Given that the force used to push the crate is F = 200N

The force directed 20° below the horizontal

Mass of crate is m = 25kg

Weight of the crate can be determine using

W = mg

g is gravitational constant =9.8m/s²

W = 25×9.8

W = 245 N

Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

ΣFy = m•ay

ay = 0, since the body is not moving in the vertical or y direction

N—W—F•Sin20 = 0

N = W+F•Sin20

N = 245+ 200Sin20

N = 245 + 68.4

N = 313.4 N

The normal force is approximately 310 N to the nearest ten

3 0

2 years ago

What type of diagram is this?

FromTheMoon [43]

Answer: That’s a sankey diagram

Explanation:

5 0

2 years ago

A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the c

Romashka [77]

Answer:

Explanation:

Given that,

Assume number of turn is

N= 1

Radius of coil is.

r = 5cm = 0.05m

Then, Area of the surface is given as

A = πr² = π × 0.05²

A = 7.85 × 10^-3 m²

Resistance of

R = 0.20 Ω

The magnetic field is a function of time

B = 0.50exp(-20t) T

Magnitude of induce current at

t = 2s

We need to find the induced emf

This induced voltage, ε can be quantified by:

ε = −NdΦ/dt

Φ = BAcosθ, but θ = 90°, they are perpendicular

So, Φ = BA

ε = −NdΦ/dt = −N d(BA) / dt

A is a constant

ε = −NA dB/dt

Then, B = 0.50exp(-20t)

So, dB/dt = 0.5 × -20 exp(-20t)

dB/dt = -10exp(-20t)

So,

ε = −NA dB/dt

ε = −NA × -10exp(-20t)

ε = 10 × NA exp(-20t)

Now from ohms law, ε = iR

So, I = ε / R

I = 10 × NA exp(-20t) / R

Substituting the values given

I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2

I = 1.67 × 10^-18 A

6 0

3 years ago

An athlete is running a 400m race around a 400m track. On the backstretch the athlete's velocity is 8m/s but he is running into

Aleksandr-060686 [28]

Answer:

33 N

Explanation:

v = Velocity of fluid = 8+2 = 10 m/s

$\rho$ = Density of fluid = 1.2 kg/m³

C = Coefficient of drag = 1.1

A = Cross sectional area = 0.5 m²

Drag force is given by

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 1.1\times 0.5\times (8+2)^2\\\Rightarrow F=33\ N$

The drag force on the athlete is 33 N

3 0

2 years ago