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2 years ago

A projectile is launched straight up. From launch to landing back down takes 8s.

1 answer:

6 0

39.2 m/s is the launch velocity.
This will make the velocity zero after 4 seconds, and then 32.9m/s after 8 seconds as it lands back down.

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24 A person stands at the side of a straight railway track. A train moves towards the person and emits sound from its whistle. T

Andrei [34K]

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6 0

3 years ago

A football is thrown horizontally with an initial velocity of (16.6 m/s)x^. ignoring air resistance, the average acceleration of

Andrei [34K]

Solution:

According to the equations for 1-D kinematics. The only change to them is that instead one equation that describes general motion.

So we will have to use the equations twice: once for motion in the x direction and another time for the y direction.

v_f=v_o + at ……..(a)

[where v_f and v_o are final velocity and initial velocity, respectively]

Now ,

Initially, there was y velocity, however gravity began to act on the football, causing it to accelerate.

Applying this value in equation (a)

v_yf = at = -9.81 m/s^s * 1.75 = -17.165 m/s in the y direction

For calculating the magnitude of the equation we have to square root the given value

(16.6i - 17.165y)

\\ \left | V \right |=sqrt{16.6^{2}+17.165^{2}}\\ = \sqrt{275.56+294.637225}\\= \sqrt{570.197225}\\= 23.87[/tex]

5 0

3 years ago

Read 2 more answers

What happens to speed when wavelength decreases

Phantasy [73]

Speed = wavelength × frequency

giving that frequency is 0, wavelength and speed are directionally proportional. wavelength decrease = speed decrease

6 0

2 years ago

What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?

MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^2}$

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

$I = \frac{100}{4\pi (2.5)^2}$

$I = 1.2738W/m^2$

The relation between intensity I and $E_{max}$

$I = \frac{E_max^2}{2\mu_0 c}$

Here,

$\mu_0$ = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

$E_{max}^2 = 2I\mu_0 c$

$E_{max}=\sqrt{2I\mu_0 c }$

$E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)$

$E_{max} = 30.982 V/m$

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

$B_{max} = \frac{E_{max}}{c}$

$B_{max} = \frac{30.982 V /m}{3*10^8}$

$B_{max} = 1.03275 *10{-7} T$

Therefore the maximum value of the magnetic field is $B_{max} = 1.03275 *10{-7} T$

3 0

3 years ago

A 70- kg bicycle rides his 9.8- kg bicycle with a speed of 16 m/ s. What is the magnitude of the braking force of the bicycle co

Rus_ich [418]

Answer:

F = -319.2 N

Explanation:

Given that,

The mass of a bicyclist, m = 70 kg

Mass of the bicycle = 9.8 kg

The speed of a bicycle, v = 16 m/s

We need to find the magnitude of the braking force of the bicycle come to rest in 4.0 m.

The braking force is given by :

$F=ma\\\\=\dfrac{m(v-u)}{t}\\\\=\dfrac{(70+9.8)(0-16)}{4}\\\\=-319.2\ N$

So, the required force is 319.2 N.

3 0

2 years ago