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3 years ago

If Karen rode her bicycle 5 km to school in 30 minutes, what was her average speed? Give your answers in km/h.

Physics

1 answer:

Doss [256]3 years ago

5 0

Answer:

10 km/h

Explanation:

Before calculating the average speed, we need to convert the time taken for the trip from minutes to hours. Since 1 hour cointains 60 minutes, we have:

$1 h: 60 min = t : 30 min\\t=\frac{1 h \cdot 30 min}{60 min}=0.5 h$

Karen's average speed is given by:

$v=\frac{d}{t}$

where

d = 5 km is the distance covered

t = 0.5 h is the time taken for the trip

Substituting into the equation, we find

$v=\frac{5 km}{0.5 h}=10 km/h$

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An 8.5 kg crate is pulled 5.1 m up a 30 degree incline by a rope angled 17 degrees above the incline. The tension in the rope is

Ulleksa [173]

Answer:

1. a $W_t=746.63 J$

b $E_p=212.415 J$

c $W_n=183.96J$

2. $T_e=99.71J$

Explanation:

a). The work done by the tension is:

$W_t=T*dt$

$dt=\frac{5.1}{cos(17)}$

$W_t=140N*\frac{5.1}{cos(17)}$

$W_t=746.63 J$

b). The work done potential of gravity

$E_p=m*g*h$

$h=5.1*sin(17)$

$E_p=8.5kh*9.8*5.1*sin(30)$

$E_p=212.415 J$

c). The work done by the normal force

$W_n=N*d_n$

$d_n=5.1*sin(30)=2.55$

$W_n=8.5kg*9.8*cos(30)*2.55$

$W_n=183.96J$

2. The increase in thermal energy is:

$T_e=F*d$

$F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)$

$F_k=19.5N$

$T_e=19.55*5.1m$

$T_e=99.71J$

4 0

3 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

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3 years ago

What is the relationship between Avogadro's number and a mole?

Shtirlitz [24]

Answer:

A. There are 6.02 x 1023 items in a mole, which equals Avogadro's

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Explanation:

The mole of a substance is

8 0

2 years ago

a circular loop is hanging on the wall. it has a radius of 33.3 cm and is comprised of 12 coils. there is a magnetic field perpe

Vika [28.1K]

Answer:

I = 2.19A, anticlockwise direction.

Explanation:

Given r = 33cm = 0.33m, N = 12, ΔB = 7.5 - 1.5 = 6.0T, Δt = 3s, R = 3.75Ω

By Faraday's law of electromagnetic induction when there is a change in flux in a coil or loop, an emf is induced in the coil or loop which is proportional to the time rate of change of the magnetic flux through the loop.

The emf E is related to the flux by the formula

E = – NdФ/dt

Where N = number of turns in the coil, Ф = magnetic flux through the loop = BA, B = magnetic field strength, A = Area

In this problem the strength of the magnetic field changes. As a result the flux too changes and an emf is induced in the coil.

ΔФ = ΔB×A = ΔB×πr² = 6×π×0.33² = 2.05Wb

E = -NΔФ/Δt = 12×2.05/3 = 8.2V

I = E/R = 8.2/ 3.75 = 2.19A

The direction of the current can be found by pointing the thumb of your right hand in the direction of the magnetic field and curling the remaining fingers around this direction. The direction of the curl of these fingers give the direction of current which in this case is anticlockwise.

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3 years ago

If you drag a sled on the ground, what does the sled experience?

Ilya [14]

The sled experiences friction against the ground

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2 years ago