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3 years ago

15

North Africa is famous for its hand-woven carpets. A-true B-false

1 answer:

Sidana [21]3 years ago

4 0

False
Step by step explanation

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A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

HACTEHA [7]

Answer:

$7.1$ ms⁻¹

Explanation:

$d$ = diameter of merry-go-round = 4 m

$r$ = radius of merry-go-round = $\frac{d}{2}$ = $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

$m$ = mass of ryan = 70 kg

Using conservation of angular momentum

$Iw_{i} - m v r = (I + mr^{2})w_{f}$

$(500)(2.0) - (70) v (2) = (I + mr^{2})(0)$

$1000 = 140 v$

$v = 7.1$ ms⁻¹

5 0

3 years ago

Which astronomer supported the belief that earth was at the center fo the universe?

MissTica

B. Ptolemy believed that the earth was the center of the universe

4 0

3 years ago

Read 2 more answers

What graph shape is this what does the snap tell you

vekshin1

The picture is hard to see but if you still need help message me

7 0

3 years ago

The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca

d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

$U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}$

the elastic potencial energy is given by:

$U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J$

The work is defined as:

$W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J$

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

$U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J$

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

$8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s$

3 0

3 years ago

When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When

Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, $\dfrac{dI}{dt}=0.0240\ A/s$

Induced emf, $\epsilon=12.4\ mV=12.4\times 10^{-3}\ V$

Current, I = 1.5 A

Magnetic flux, $\phi=0.00338\ Wb$

The induced emf through the solenoid is given by :

$\epsilon=L\dfrac{dI}{dt}$

or

$L=\dfrac{\epsilon}{(di/dt)}$ ........(1)

The self inductance of the solenoid is given by :

$L=\dfrac{N\phi}{I}$ .........(2)

From equation (1) and (2) we get :

$\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}$

N is the number of turns in the solenoid

$N=\dfrac{\epsilon I}{\phi (dI/dt)}$

$N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}$

N = 229.28 turns

or

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

3 0

3 years ago