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4 years ago

Weight measures:

Physics

1 answer:

GrogVix [38]4 years ago

3 0

C. Gravity's effect on an object. Weight = the mass of your object X the acceleration of gravity (9.8).

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What is the volume of water in 150ml of the 35% w/w of sucrose solution with a specific gravity of 1.115?

pentagon [3]

Answer:108.71 mL

Explanation:

Given

Volume of sample V=150 mL

concentration of sucrose solution 35 % w/w i.e. In 100 gm of sample 35 gm is sucrose

specific gravity =1.115

Density of solution $\rho _s=1.115\times density\ of\ water$

Thus

$\rho _s=1.15\times 1 gm/mL=1.115\ gm/mL$

mass of sample $M=1.115\times 150=167.25\ gm$

mass of sucrose $m_s=0.35\times 167.25=58.53\ gm$

mass of Water $m_w=108.71 gm$

Volume of water $=108.71\times 1=108.781 mL$

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3 years ago

An example of unprofessional behavior is?

nika2105 [10]

B, gossiping is unproessional

7 0

3 years ago

Read 2 more answers

The diameter of a 12-gauge copper wire is 0.081 in. The maximum safe current it can 17) carry (in order to prevent fire danger i

Soloha48 [4]

Answer:

0.44m/s

Explanation:

drift velocity=I/nAq

diameter 12 gauge

wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*

V = 0.44m/s

6 0

4 years ago

Hi, pls help with this question. Thank you.

topjm [15]

Answer:

i: Gamma Rays

ii: Infrared

Explanation:

It just is:)

5 0

4 years ago

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a

Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

$d = v*t \rightarrow v = \frac{d}{t}$

<u>Where</u>:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

$t = \sqrt{\frac{2d}{g}}$

<u>Where:</u>

g: is gravity = 9.8 m/s²

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s$

Now, the horizontal velocity of the rock is:

$v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s$

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.

2. To calculate the distance at which the projectile will land, first, we need to find the time:

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s$

So, the distance is:

$d = v*t = 6.67 m/s*1.43 s = 9.54 m$

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!

7 0

3 years ago