All categories

3 years ago

NEED HELP!!!! 11 POINTS!!!!

Physics

2 answers:

Ede4ka [16]3 years ago

6 0

Answer:

Explanation:

,....,.....,...,...,....,.....,......

Anna007 [38]3 years ago

5 0

Answer:

Explanation:

You might be interested in

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

a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f

34kurt

Answer:

$\frac{h_{liquid} }{ h_{body} }$ = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

B = ρ_liquid g V_liquid

let's write the translational equilibrium condition

B - W = 0

let's use the definition of density

ρ_body = m / V_body

m = ρ_body V_body

W = ρ_body V_body g

we substitute

ρ_liquid g V_liquid = ρ_body g V_body

$\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }$

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

V = A h_bogy

Thus

$\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }$

we substitute

5/9 = $\frac{h_{liquid} }{ h_{body} }$

8 0

3 years ago

Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect

Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

Where Vo = initial voltage

e = natural logarithm = 2.718

For the first capacitor 2.50 µF,

V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

4 0

3 years ago

The acceleration due to gravity on the surface of Mars is gmars = 3.7 m/s2. How much would an 11 kg bag of potatoes weigh on Mar

allsm [11]

Fnet = (mass) (acceleration)
= 11 kg x 3.7m/s^2
= 41 N

6 0

4 years ago

Find the volume of a sphere of radius 10 mm.

vredina [299]

Answer:

Explanation: This is done using the equation:

$\frac{4}{3} π R^{3}$

Because the Radius is a know value. We have the following.

$\frac{4}{3} π (10mm)^{3}$

Which is:

4188.7902 mm

5 0

3 years ago