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Contact [7]
2 years ago
10

The weight of the atmosphere above 1 m- of

Physics
1 answer:
mars1129 [50]2 years ago
4 0

Answer:

1.09 kg.m

Explanation:

no need

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An 80.0 kg hiker walks a distance of 400.0 m along a road that slopes 5.0 degrees upward, and then stops. What is the hiker's fi
lana66690 [7]
The height difference is found by
\delta H=400sin(5 \°)=34.86m
Then the change in potential energy is
E=mgh=(80.0kg)(9.8 \frac{kgm}{s^2})(34.86)= 27332J
4 0
3 years ago
Fill in the blanks. When the northern hemisphere experiences _________, the southern hemisphere experiences __________.
schepotkina [342]

I choose the letter B

8 0
3 years ago
A football player with a mass of 88 kg and a speed of 2.0 m/s collides head-on with a player from the opposing team whose mass i
Ket [755]

Answer:

Speed of another player, v₂ = 1.47 m/s

Explanation:

It is given that,

Mass of football player, m₁ = 88 kg

Speed of player, v₁ = 2 m/s

Mass of player of opposing team, m₂ = 120 kg

The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :

m_1v_1+m_2v_2=(m_1+m_2)V

V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.

v_2=-\dfrac{m_1v_1}{m_2}

v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}

v_2=-1.47\ m/s

So, the speed of another player is 1.47 m/s. Hence, this is the required solution.

7 0
3 years ago
In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this
soldier1979 [14.2K]

Answer:

1). v = - 2960526m/s

2). Toward us

3). v = - 493421m/s

4). Toward us

5). v = 1480263m/s

6).  Away from us

7). v = 3207236m/s

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm (\lambda_{0} = 121.6nm)

Redshift: \lambda_{measured} > \lambda_{0}

Blueshift: \lambda_{measured} < \lambda_{0}

Then, for this particular case it is gotten:

Star 1: \lambda_{measured} = 120.4nm

Star 2: \lambda_{measured} = 121.4nm

Star 3: \lambda_{measured} = 122.2nm

Star 4: \lambda_{measured} = 122.9nm

Star 1:

Blueshift: 120.4nm < 121.6nm

Toward us

Star 2:

Blueshift: 121.4nm < 121.6nm

Toward us

Star 3:

Redshift: 122.2nm > 121.6nm

Away from us

Star 4:

Redshift: 122.9nm > 121.6nm

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (1)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}}) (2)

<em>Case for star 1 \lambda_{measured} = 120.4 nm:</em>

<em></em>

v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})

v = - 2960526m/s

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 \lambda_{measured} = 121.4 nm:</em>

v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})

v = - 493421m/s

<em>Case for star 3 \lambda_{measured} = 122.2 nm:</em>

v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})

v = 1480263m/s

<em>Case for star 4 \lambda_{measured} = 122.9 nm:</em>

v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})

v = 3207236m/s

4 0
3 years ago
A crate on a motorized cart starts from rest and moves with a constant eastward acceleration ofa= 2.60 m/s^2. A worker assists t
Leona [35]

Answer:

To obtain the power, we first need to find the work made by the force.

1) To calculate the work, we need the next equation:

\int\limits {F} \, dx

So the force is given by the problem so our mission is to find 'dx' in terms of 't'

2) we know that:

\frac{dV}{dt} = a = 2.6

So we have:

v = 2.6t

Then:

\frac{dx}{dt} = V = 2.6t\\ \\dx = 2.6t*dt

3) Finally, we replace everything:

\int\limits^{4.7}_{0} {5.4t*2.6t} \, dt

After some calculation, we have as a result that the work is:

161.9638 J.

4) To calculate the power we need the next equation:

P = \frac{W}{t}

So

P = 161.9638/4.7 = 34.46 W

8 0
3 years ago
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