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kompoz [17]
3 years ago
5

A horse pulls forward on a carriage with a given force. By Newton's Third Law, the carriage must be pulling on the horse backwar

d with an equal and opposite force. Given this, what explains why the horse and carriage can move forward? Two brown horses pulling wooden carriage Image from Wikimedia Commons, CC BY-SA 2.0 Choose 1 answer: Choose 1 answer: A The cart is rolling on wheels while the horse's hooves have traction with the ground B The forward force of the horse is just big enough to overcome the backward force of the cart and start the cart forward C The cart's force is only in reaction to the horse's force so it does not define direction of movement D There is a brief moment where the horse pulls before the reaction force kicks in E The forward and backward forces are equal, so it actually can't move forward
Physics
1 answer:
olga2289 [7]3 years ago
6 0
The cart's force is only in reaction to the horse's force so it does not define direction of movement.
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You want to create a new warming appliance that is safe. Which type of electromagnetic wave would be most useful to investigate?
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Infrared waves. The other 3 would hurt you because they are below the UV scale of 400 nanometers. Infrared is light that wouldn't harm you
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A 15.0 kg turntable with a radius of 25 cm is covered with a uniform layer of dry ice that has a mass of 9.0 kg. The angular spe
liubo4ka [24]

Answer:

 ω₂=1.20

Explanation:

Given that

mass of the turn table ,M= 15 kg

mass of the ice ,m= 9 kg

radius ,r= 25 cm

Initial angular speed ,ω₁ = 0.75 rad/s

Initial mass moment of inertia

I_1=\dfrac{M+m}{2}r^2

I_1=\dfrac{15+9}{2}\times 0.25^2\ kg.m^2

I_1=0.75\ kg.m^2

Final mass moment of inertia

I_2=\dfrac{M}{2}r^2

I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2

I_2=0.468\ kg.m^2

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

\omega_2=\dfrac{0.75\times 0.75}{0.468}\ rad/s

ω₂=1.20

7 0
3 years ago
A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring
sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0
3 years ago
How much have jellyfish populated since 2015?
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About a mil sience 2014-2015
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4 years ago
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If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w
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Answer:

P(3600)=593.247W

Explanation:

First, let's find the voltage through the resistor using ohm's law:

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AC power as function of time can be calculated as:

P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)  (1)

Where:

\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency

Because of the problem doesn't give us additional information, let's assume:

\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W

5 0
3 years ago
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