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3 years ago

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A person holding a lunch bag is moving upward in a hot air balloon at a constant speed of 7.3 m/s . When the balloon is 24 m abo

ve the ground, she accidentally releases the bag. What is the speed of the bag just before it reaches the ground?

1 answer:

Kitty [74]3 years ago

8 0

Explanation:

Given that,

Initial speed of the bag, u = 7.3 m/s

Height above ground, s = 24 m

We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :

$v^2=u^2+2as$

$v^2=(7.3)^2+2\times 9.8\times 24$

$v=\sqrt{523.69}$

v = 22.88 m/s

So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.

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What is the most abundant gas in the atmosphere?

Snowcat [4.5K]

Answer: Nitrogen

Explanation: Nitrogen makes up 78% of the atmosphere, Oxygen makes up 21%, and Argon 0.9%.

Water vaper makes up between 1-4%, depending on the region.

Carbon Dioxide makes up only about 0.04%.

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3 years ago

A. If a T-Rex runs 20 meters in 4 seconds, what is it’s average speed?

mariarad [96]

5m/s

100m

Explanation:

Average speed is sum of distance distance traveled in a given time by a body.

Average speed= $\frac{total distance covered}{time taken}$

Distance = 20m

time = 4s

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For the spaceship;

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Speed = 50m/s

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learn more:

Average speed brainly.com/question/5063905

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3 years ago

The wavelength of the visible line in the hydrogen spectrum that corresponds to m = 5 in the Balmer equation is: A. 656 nm. B. 4

BaLLatris [955]

Answer:

The wavelength of the visible line in the hydrogen spectrum is 434 nm.

Explanation:

It is given that, the wavelength of the visible line in the hydrogen spectrum that corresponds to n₂ = 5 in the Balmer equation.

For Balmer series, the wave number is given by :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

R is the Rydberg's constant

For Balmer series, n₁ = 2. So,

$\dfrac{1}{\lambda}=1.097\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{5^2})$

$\lambda=4.34\times 10^{-7}\ m$

or

$\lambda=434\ nm$

So, the wavelength of the visible line in the hydrogen spectrum is 434 nm. Hence, this is the required solution.

6 0

3 years ago

A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.7 m/s^2.

Komok [63]

Answer:

The coefficient of static friction between the car and the track

u=0.572

Explanation:

We don't know the mass of the car or any other information so the acceleration is the reason to solve the friction coefficient

∑ $F=F_{f}=m*a_{t}$

As we know

$F_{f}=u*F_{N}=u*m*g$

Also the center ward direction forces

$F_{fc}=m*a_{c}$

$a_{c}=\frac{v_{t}^2}{r}$

$F_{fc}=m*\frac{v_{t}^2}{r}$

But now vt relation with the tangential acceleration

$v_{t}=2*a_{t}*\frac{\pi }{r}$

replacing

$F_{fc}=m*a_{t}*\frac{2\pi*r}{2r}$

$F_{fc}=m*a_{t}*\pi$

So magnitude of the force can get by

$F_{f}=\sqrt{(m*a_{t}*\pi)^{2}+(m*a_{t})^{2}}$

Get the factor to simplify

$F_{f}=a_{t}*m*\sqrt{(1+\pi^2)}$

$u*m*g=m*a_{t}*(\sqrt{1+\pi^2})$

Solve to u'

$u=\frac{a_{t}}{g}*(\sqrt{1+\pi^2})$

$u=\frac{17\frac{m}{s^2} }{9.8\frac{m}{s^2}}*(\sqrt{1+\pi^2})=0.572$

5 0

3 years ago

A circuit is supplied with 30 volts and the total circuit resistance is 3 kΩ. What's the current flowing through the circuit?

Daniel [21]

Answer: The current flowing through the circuit is 0.01A (or 10 mA)

Explanation:

Use Ohm's Law:

$I=\frac{U}{R}$

Given the values of U=30V and R=3000Ohm:

$I=\frac{30 V}{3000 \Omega}=0.01A=10mA$

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3 years ago