1. What was the electromagnetic spectrum ?

the pull of gravity on mars is 3.7m/s^2. if a astronaut on mars lifts a 10 kg rock 1 m off the ground, just to see whats under i

Elanso [62]

Gravitational potential energy can be calculated using the formula:

$PE_{grav} =mgh$

Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height

On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.

So put in what you know and solve for what you don't know.
m = 10kg
g = 3.7m/s^2
h = 1m

So we put that in and solve it.
$PE_{grav} =mgh$
$PE_{grav} =(10kg)(3.7m/s^{2})(1m)$
$PE_{grav} =37J$

7 0

4 years ago

An astronaut weighs 863 N on Earth. What is the astronaut’s mass?

Alecsey [184]

86.3 pounds.... I think

6 0

4 years ago

Read 2 more answers

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.04 m away from a waterfall 0.585 m in heigh

S_A_V [24]

Answer:

V₀ = 5.47 m/s

Explanation:

The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 3.04 m

θ = Launch Angle = 41.7°

V₀ = Minimum Launch Speed = ?

g = 9.81 m/s²

Therefore,

3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)

V₀² = 3.04 m/(0.10126 s²/m)

V₀ = √30.02 m²/s²

6 0

3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago

To find the number of neutrons in an atom, what would you subtract?

kolezko [41]

You look up the element in the periodic table. Subtract the atomic number (small number) from the mass number (big number). The answer is the number of neutrons.

7 0

3 years ago