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GrogVix [38]
2 years ago
7

please help me with my question I will like and mark as brainliest NO LINKS THEY DON'T WORK AND IF U DON'T KNOW THE ANSWER PLS D

ON'T ANSWER AT ALL​

Physics
1 answer:
bearhunter [10]2 years ago
7 0

Answer:

a)

Weight in Air = 0.3N

Weight in Water = 0.25N

Weight in Liquid = 0.24N.

Upthrust /Buoyant Force = Weight in Air – Weight in Fluid(Water in this case)

= 0.3 – 0.25

= 0.5N.

b) R.D of Body = Density of Body/Density of Standard Fluid(Water).

There's a Derived Formula for RD.

I'm gonna Apply it here.

Ask me for the derivation in the Comment section if you need it.

RD = α/ρ = (Weight in Air) / (Upthrust Force)

Where

α = density of the Body(or reference substance)

ρ = density of standard fluid (water)

= 0.3/0.05 = 6.

c) RD of Liquid = (Density of Liquid) /(Density of standard Fluid(water)

Or we just go by that formula

RD of Liquid = Weight in Air/Upthrust(In Liquid)

We'll be using the Upthrust in that Liquid now.

= 0.3 – 0.24 = 0.06

RD = 0.3/0.06 = 5.

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NASA’s Tracking and Data Relay Satellite (TDRS) System constellation resides at geosynchronous orbit (35,000km) altitude. If a t
Leya [2.2K]

Answer:

35,000 km = 35,000,000 m = 3.5 E107 m

t = S / v = 3.5 * 10E7 / 3.0 E10E8 = .117 sec

6 0
2 years ago
A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr
fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}

The values are

\displaystyle y_o=15\ m

\displaystyle v_o=14\ m/s

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

\displaystyle y=0

\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0

Multiplying by 2

\displaystyle 2y_o+2v_ot-gt^2=0

Clearing the coefficient of t^2

\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0

Plugging in the given values, we reach to a second-degree equation

\displaystyle t^2-2.857t-3.061=0

The equation has two roots, but we only keep the positive root

\displaystyle \boxed {t=3.69\ s}

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

\displaystyle Y_r=15+V_r.t

\displaystyle Y_r=15+14\times3.69

\displaystyle \boxed{Y_r=66.62\ m}

3 0
2 years ago
The diver on the diving board is 10 meters high and has a mass of 0.050kg, what is his GPE?
sveta [45]

Answer:

<h2>4.9 J</h2>

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 10 × 9.8 × 0.05

We have the final answer as

<h3>4.9 J</h3>

Hope this helps you

6 0
3 years ago
An 800 N box is pushed across a level floor for a distance of 5.0 m with a force of 400 N. How much work was done on this box.
AveGali [126]

Answer: 2000 J.

Explanation: Since work is force*displacement, we just have to multiply the force by the distance: w = f*d = 400 N*5.0 m = 2000 J.

4 0
2 years ago
qual o nome da posição que o planeta tem maior velocidade? este ponto é mais próximo ou mais afastado do sol?
klio [65]

Em inglês, esse ponto é 'perihelion' ... 'periélio'. É o mais próximo que a Terra ou o planeta chega do sol.

7 0
3 years ago
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