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inna [77]
2 years ago
4

a woman throws her keys straight up at 8.49 m/s. her friend catches them 2.89 m higher. how long were the keys in the air? (unit

= s)​
Physics
1 answer:
romanna [79]2 years ago
5 0
Alright this one is just kinematics, you can use one of the big four equations and in this case this one works because acceleration is constant and you don’t have the final velocity to worry about:
Δx= displacement
Vi= initial velocity
T= time
A= acceleration

Δx=(Vi)(t)+(1/2)(a)(t^2)

A=9.8m/s2
Vi=8.49 m/s
Δx=2.89 meters

Then plug in:

2.89=(8.49)(t)+(1/2)(9.8)(t^2)

Then solve for t and you’re all set.
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Consider a person standing in an elevator that is moving at constant speed upward. The person, of mass m, has two forces acting
larisa [96]

Answer:

The weight of the person has a smaller magnitude.

Explanation:

For an observer in inertial frame of reference for the person in the elevator Newton's Second Law can be written as

Normal reaction acts upwards

Weight acts downwards

\sum F_{v}=ma_{v}\\\\N-mg=m\times a_{v}\\\\m\times a_{v}> 0\\\\\therefore N-mg> 0\\\\\therefore N> mg

Here

N is the normal reaction force

mg is the weight of the person

g is acceleration due to gravity

4 0
2 years ago
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A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a
pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

120^2=130^2+50^2-2\times130\times50\cos\alpha

\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}

\alpha=\cos^{-1}(0.3846)

\alpha=67.38^{\circ}

We need to calculate the angle β

Using cosine law

50^2=130^2+120^2-2\times130\times120\cos\beta

\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}

\beta=\cos^{-1}(0.923)

\beta=22.63^{\circ}

We need to calculate the force on 130 cm side

Using formula of force

F_{130}=ILB\sin\theta

F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0

F_{130}=0

We need to calculate the force on 120 cm side

Using formula of force

F_{120}=ILB\sin\beta

F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63

F_{120}=0.1385\ N

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

F_{50}=ILB\sin\alpha

F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38

F_{50}=0.1385\ N

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

8 0
2 years ago
In this lab you will use a cart and a track to explore Newton's second law of motion. You will vary two different variables and
Savatey [412]

<u>Answer:</u>

<em>Newtons II law: </em>

<em>     </em>It  is defined as<em> "the net force acting on the object is a product of mass and acceleration of the body"</em> . Also it defines that the <em>"acceleration of an object is dependent on net force and mass of the body".</em>

Let us assume that,a string is attached to the cart, which passes over a pulley along the track. At another end of the string a weight is attached which hangs over the pulley. The hanging weight provides tension in the spring, and it helps in accelerating the cart. We assume that the string is massless and no friction between pulley and the string.

Whenever the hanging weight moves downwards, the cart will accelerate to right side.

<em>For the hanging weight/mass</em>

When hanging weight of mass is m₁ and accelerate due to gravitational force g.

           Therefore we can write F = m₁ .g

and the tension acts in upward direction T (negetive)

        Now, Fnet = m₁ .g - T

                          = m₁.a

So From Newtons II law<em> F =  m.a</em>

3 0
2 years ago
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A 1-m-long monopole car radio antenna operates in the AM frequency of 1.5 MHz. How muchcurrent is required to transmit 4 W of po
Zanzabum

Answer:

The current needed to transmit Power of 4 W is 28.47 A

Solution:

As per the question:

Length of the antenna, L_{a} = 1 m

Frequency, \vartheta = 1.5 MHz = 1.5\times 10^{6} Hz

Power transmitted, P_{t} = 4 W

Now,

For a monopole antenna:

\lambda_{a} = \frac{c}{\vartheta}

where

\lambda_{a} = wavelength transmitted by the antenna

c = speed of light in vacuum

\lambda_{a} = \frac{3\times 10^{8}}{1.5\times 10^{6}} = 200 m

Now,

Since, the value of \lambda_{a} >> L_{a} thus the monopole is a Hertian monopole.

The resistance is calculated as:

R = \frac{1}{2}(\frac{dL_{a}}{\lambda_{a}})^{2}\times 80\pi^{2}

R = \frac{1}{2}(\frac{1}{200)^{2}\times 80\pi^{2} = 9.869\times 10^{- 3} = 9.869 m\Omega

P_{radiated} = P_{t}

P_{radiated} = \frac{R}{I^{2}}

Now, the current I is given by:

I = \sqrt{\frac{2P_{t}}{R}} = \sqrt{\frac{2\times 4}{9.869\times 10^{- 3}}} = 28.47 A

5 0
2 years ago
According to the nebular theory of solar system formation, what key difference in their early formation explains why the jovian
svp [43]

Answer:

The Jovian planets formed beyond the Frostline while the terrestrial planets formed in the Frostline in the solar nebular

Explanation:

The Jovian planets are the large planets namely Saturn, Jupiter, Uranus, and Neptune. The terrestrial planets include the Earth, Mercury, Mars, and Venus. According to the nebular theory of solar system formation, the terrestrial planets were formed from silicates and metals. They also had high boiling points which made it possible for them to be located very close to the sun.

The Jovian planets formed beyond the Frostline. This is an area that can support the planets that were made up of icy elements. The large size of the Jovian planets is as a result of the fact that the icy elements were more in number than the metal components of the terrestrial planets.

3 0
2 years ago
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