How much energy is needed to melt 600 g of ice at 0 degrees C?

Which of the following best describes what went wrong with the scientists’ study? an improper experimental procedure the lack of

Rasek [7]

B- lack of control.

the scientists did not have a proper control group.

3 0

3 years ago

Read 2 more answers

A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid

Galina-37 [17]

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm= $\frac{6}{100}=0.06 m$

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

$dF=IBdr$

$\tau=rdF=IBrdr$

$\tau=\int_{0}^{R}IBr dr$

$\tau=IB(\frac{R^2}{2}$

Torque due to friction

$\tau=R\times F$

Where friction force=F

$R\times F=\frac{IBR^2}{2}$

$F=\frac{IBR}{2}$

Substitute the values

$F=\frac{3\times 1\times 0.06}{2}$

$F=0.09 N$

7 0

3 years ago

A 25 kg child runs at a speed of 5.0 m/s and jumps onto a stationary shopping cart and holds on for dear life. The cart has mass

makkiz [27]

Answer:

3.38 m/s

Explanation:

Mass of child = m₁ = 25

Initial speed of child = u₁ = 5 m/s

Initial speed of cart = u₂ = 0 m/s

Mass of cart = m₂ = 12 kg

Velocity of cart with child on top = v

This is a case of perfectly inelastic collision

$m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\frac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\frac{25\times 5+12\times 0}{25+12}\\\Rightarrow v=\frac{125}{37}\\\Rightarrow v=3.38\ m/s$

Velocity of cart with child on top is 3.38 m/s

7 0

3 years ago

A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a

Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0

3 years ago

The space shuttle usually orbited Earth at altitudes of around 300.0 km. 1) Determine the time for one orbit of the shuttle abou

777dan777 [17]

Answer:

Explanation:

distance of shuttle from centre of the earth = radius of the orbit

= 6300 + 300 = 6600 km

= 6600 x 10³

Formula of time period of the satellite

T = 2π R /v₀ , v₀ is orbital velocity

v₀ = √gR , ( if height is small with respect to radius )

T = 2π R /√gR

= 2π√ R /√g

= 2 x 3.14 x √ 6600 x 10³ / √9.8

= 2 x 3.14 x 256.9 x 10 / 3.13

= 5154.41 s

= 5154.41 / 60 minutes

= 85.91 m

85.9 minutes.

2 ) No of sunrise per day = no of rotation per day

= 24 x 60 / 85.9

= 16.76

or 17 sunrises.

3 0

3 years ago