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

An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same

Hatshy [7]

Answer:

$v_2=\sqrt{2}v_1$

Explanation:

The velocity v₁ can be calculated with the kinematic formula:

$v_1^{2} =v_0^{2} +2gh$

Since the object is initially at rest, v₁ becomes:

$v_1=\sqrt{2gh}$

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

$v_2^{2}=v_1^{2} +2gh$

Substituting v₁ in this expression and solving for v₂, we get:

$v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}$

Now, dividing v₂ over v₁, we get the expression:

$\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\ \\\implies v_2=\sqrt{2}v_1$

It means that v₂ is √2 times v₁.

4 0

3 years ago

Net force causes motion

ASHA 777 [7]

Answer:

yes

Explanation:

8 0

2 years ago

Read 2 more answers

A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration?

Scrat [10]

Answer:

${ \boxed{ \bold{ \sf{Acceleration \: ( \: a) = 8 \: m/ {s \: }^{2} }}}}$

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♨ Question :

A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration ?

♨ $\underbrace{ \sf{Required \: Answer : }}$

☄ Given :

Initial velocity ( u ) = 0
Final velocity ( v ) = 60.0 m/s
Time ( t ) = 7.50 s

☄ To find :

Acceleration ( a )

✒ We know ,

$\boxed{ \underline{ \bold{ \sf{Acceleration \: ( \: a) = \frac{Final velocity ( v ) - Initial velocity ( u)}{t} }}}}$

Substitute the values and solve for a.

➛ $\sf{a = \frac{60.0 - 0}{7.50}}$

➛ $\sf{a = \frac{60.0}{7.50}}$

➛ $\boxed{ \boxed{ \sf{a = 8 \: m/ {s \: }^{2} }}}$

---------------------------------------------------------------

✑ Additional Info :

When a certain object comes in motion from rest , in the case , initial velocity ( u ) = 0
When a moving object comes in rest , in the case , final velocity ( v ) = 0
If the object is moving with uniform velocity , in the case , u = v.
If any object is thrown vertically upwards in the case , a = -g
When an object is falling from certain height , in the case , final velocity at maximum height ( v ) = 0.

Hope I helped!

Have a wonderful time ツ

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4 0

2 years ago

What is the current in mA through a 500 Ω resistor that is connected to a 1.5 V battery? Show all calculations using Ohm’s law (

Ludmilka [50]

Answer:

3 mA.

Explanation:

The following data were obtained from the question:

Resistor (R) = 500 Ω

Potential difference (V) = 1.5 V

Current (I) =.?

Using the ohm's law equation, we can obtain the current as follow:

V = IR

1.5 = I x 500

Divide both side by 500

I = 1.5 / 500

I = 3×10¯³ A.

Therefore, the current in the circuit is 3×10¯³ A.

Finally, we shall convert 3×10¯³ A to milliampere (mA).

This can be obtained as follow:

Recall:

1 A = 1000 mA

Therefore,

3×10¯³ A = 3×10¯³ × 1000 = 3 mA

Therefore, 3×10¯³ A is equivalent to 3 mA.

Thus, the current in mA flowing through the circuit is 3 mA.

3 0

3 years ago

Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450K, 350kPa, and a velocity of 3

MissTica

Answer:

(a) $A_1=0.283m^2$

(b) $Q=-105.5kW$ : From the air to the surroundings.

Explanation:

Hello,

(a) In this case, we can compute the area at the entrance by firstly computing the inlet volumetric flow:

$V_1=\frac{mRT_1}{P_1M}= \frac{2.3\frac{kg}{s} *8.314\frac{kPa*m^3}{kmol* K}*450K}{350kPa*28.97\frac{kg}{kmol} } =0.849\frac{m^3}{s}$

Then, with the velocity, we compute the area:

$A_1=\frac{V_1}{v_1}=\frac{0.849\frac{m^3}{s} }{3\frac{m}{s} } =0.283m^2$

(b) In this case, via the following energy balance for the nozzle:

$Q-W=H_2-H_1+\frac{1}{2} mV_2^2-\frac{1}{2} mV_1^2$

We can easily compute the change in the enthalpy by using the given Cp and neglecting the work (no done work):

$\Delta H=H_2-H_1=mCp\Delta T=2.3kg/s*1.011\frac{kJ}{kg*K}*(300K-450K)\\ \\\Delta H=-348.795kW$

Finally, the heat turns out:

$Q=-348.795kW+\frac{1}{2}*2.3\frac{kg}{s}*[(460\frac{m}{s})^2 -(3\frac{m}{s})^2 ]\\\\Q=-348.795kW+243329.65W*\frac{1kW}{1000W}\\ \\Q=-105.5kW$

Such sign, means the heat is being transferred from the air to the surroundings.

Regards.

4 0

3 years ago