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3 years ago

Compared to its mass on earth, the mass of a 10-kg object on the moon is

1 answer:

3 0

The mass are the same because mass doesnt change but its weight change because the gravity change in planets.
Hope this helps

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I need help on all of this

wel

dude is -2m/s, ...ettte is +2m/s

pos vel is when dudette, eg, is going in increasing x, in this case.

neg vel is when dudette, eg, is going in decreasing x, in this case. ie she turns round and runs

zero vel is zero speed. dudette standing still

positive vel neg pos top dia

yes

all yes looks like ...

4 0

3 years ago

A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is

puteri [66]

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

$\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}$

$\dfrac{1}{u}=\dfrac{103}{4080}$

$u =\dfrac{4080}{103}$

The magnification is

$m = \dfrac{-v}{u}$

$m=\dfrac{-240\times103}{4080}$

$m = -6.05$

Hence, The magnification is -6.05.

6 0

3 years ago

2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th

Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = $5.33$ × $10^{-5}$ T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ $B = \frac{\mu_{0}i }{2\pi R}$

Where $B =$ magnetic field due to long wires, $\mu_{0} =$ $4\pi \times10^{-7}$ , $R =$ perpendicular distance from wire to given point

From any one wire $R_{1} =$ 5 cm, $R_{2} =$ 3 cm

so we write,

∴ $B = B_{1} + B_{2}$

$B = \frac{\mu_{0} i}{2\pi R_{1} } + \frac{\mu_{0} i}{2\pi R_{2} }$

$B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]$

$B = 5.33\times10^{-5} T$

Therefore, the magnitude of magnetic field at given point = $5.33\times10^{-5} T$

3 0

3 years ago

Despite a very strong wind, a tennis player

Gnoma [55]

Answer:

Option 5. 1 and 3

Solution:

The only forces acting on the tennis ball after it has left contact with the racquet and the instant before it touches the ground are the force of gravity in the downward direction and the force by the air exerted on the ball.

The ball after it left follows the path of trajectory and as it moves forward in the horizontal direction the force of the air acts on it.

In the whole projectile motion of the ball, the acceleration due to gravity acts on the ball thus the force of gravity acts on the ball in the downward direction before it hits the ground.

6 0

3 years ago

Hi..Please answer questions???

Nadya [2.5K]

Answer:

Explanation:

Becuse its complete number

3 0

3 years ago