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RUDIKE [14]
2 years ago
6

Q2. Point D is located at 49.3 m above the ground. The speed of the roller

Physics
1 answer:
dlinn [17]2 years ago
4 0
B hope this help u!!!
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A ball is released from a hot air balloon moving downward with a velocity of -10.0 meters/second and a height of 1,000 meters. H
Nikolay [14]
Here, ball is released... and it is in free fall means with zero initial velocity.

We know, s = ut + 1/2 at²
Here, s = 1000 m
u = 0
a = 10 m/s2

Substitute their values, 
1000 = 0 + 1/2 * 10 * t²
2000 = 10 * t²
t² = 2000 /10
t = √200
t = 14.14 s

In short, Your Answer would be 14.14 seconds

Hope this helps!
7 0
3 years ago
Read 2 more answers
When a light bulb is connected to a 4.5 V battery, a current of 0.16 A passes through the bulb filament. What is the resistance
larisa [96]

Answer:

R = 28.125 ohms

Explanation:

Given that,

The voltage of a bulb, V = 4.5 V

Current, I = 0.16 A

We need to find the resistance of the filament. Using Ohm's law,

V = IR

Where

R is the resistance of the filament

So,

R=\dfrac{V}{I}\\\\R=\dfrac{4.5}{0.16}\\\\R=28.125\ \Omega

So, the resistance of the filament is equal to 28.125 ohms.

5 0
2 years ago
If the frequency of a periodic wave is doubled, the period of the wave will be
Lunna [17]
The period of the wave would be halved 
5 0
3 years ago
Please help! 30Points!! An ultrasound machine uses waves to create images. The machine uses sound waves and which wave interacti
Gnesinka [82]
The correct answer should be C
5 0
3 years ago
Read 2 more answers
Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as
viva [34]

Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

Explanation:

It is angular momentum given by

      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

Let's apply this relationship to our case

Let's start by breaking down the speed

      v₀ₓ = v₀ cosn 45

      voy =v₀ sin 45

      v₀ₓ = 9 cos 45

      voy = 9 without 45

      v₀ₓ = 6.36 m / s

      voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

   y = voy² / 2g

   y = (6.36)²/2 9.8

   y = 2.06 m

Let's calculate the angular momentum

L= \left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

5 0
3 years ago
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