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3 years ago

PLEASE HELPPP ASAP, I'LL GIVE BRAINLEST

Physics

1 answer:

zmey [24]3 years ago

6 0

Answer:

a) Joules

b) W (symbol for work) (J is symbol for Joules)

c) seconds

d) watts

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A block with mass M = 3 kg is moving on a flat surface with constant speed v1 =

Alchen [17]

Answer:

this makes no since so i cant help you here sorry

5 0

2 years ago

A wire loop of radius 0.50 m lies so that an external magnetic field of magnitude 0.40 T is perpendicular to the loop. The field

vazorg [7]

The magnitude of the induced emf is given by:

ℰ = |Δφ/Δt|

ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time

The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:

φ = BA

B = magnetic field strength, A = loop area

The area of the loop A is given by:

A = πr²

r = loop radius

Make a substitution:

φ = B2πr²

Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:

Δφ = ΔB2πr²

ΔB = change in magnetic field strength

Make another substitution:

ℰ = |ΔB2πr²/Δt|

Given values:

ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s

Plug in and solve for ℰ:

ℰ = |(-0.20)(2π)(0.50)²/2.5|

ℰ = 0.13V

3 0

3 years ago

Find the sum 5.24 g, 43.261 g, and 7.3458 g. Write your answer with the correct amount of significant figures.

lapo4ka [179]

Answer:

This is how I do it:

5.24 rounded to one significant figure is 5
43.261 rounded to one significant figure is 40.
7.3458 rounded to one significant figure is 7.
5 + 40 + 7 is 52 g

Hope this helps you.

Explanation:

7 0

3 years ago

The subatomic particle that has the less mass is the

e-lub [12.9K]

The electron has the least mass

6 0

3 years ago

A 140 g mass is connected to a light spring of force constant 5 N/m that is free to oscillate on a horizontal, frictionless trac

photoshop1234 [79]

Answer:

Explanation:

mass attached m = .14 kg

force constant k = 5N / m

displacement

= amplitude of oscillation

A = .03 m

A ) period of motion = $2\pi\sqrt{\frac{m}{k} }$

= 2 x 3.14 $\sqrt{\frac{.14}{5} }$

T = 1.05 s

B ) maximum speed of block = angular velocity x amplitude

= (2π /T) x A

= (2 x 3.14 x .03) / 1.05

= .1794 m / s

17.94 cm /s

C )

maximum acceleration = angular velocity² x amplitude

= (2π /T)² x A

= (2π /1.05)² x .03

= 1.073 m / s²

D )

position

S = A cos ωt , ω is angular velocity

S = .03 cos(2πt /T)

= .03 cos 5.98 t

v =ω A sin(2πt /T)

= 5.98 x .03 sin5.98t

= .1794 sin5.98t

acceleration = ω²A sin5.98t

= 1.073 sin5.98t

7 0

3 years ago

PLEASE HELPPP ASAP, I'LL GIVE BRAINLEST​

PLEASE HELPPP ASAP, I'LL GIVE BRAINLEST