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3 years ago

You are on earth, and have an object with a mass of 7 kg. with how much force does the earth's gravity pull the object downward?

Physics

1 answer:

svp [43]3 years ago

6 0

This is equivalent to asking the mass’s weight. Force=mass * acceleration. So 7*9.8= 68.6 N

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A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wave

Ann [662]

Answer:

$\lambda$ = 40 cm

Explanation:

given data

string length = 30 cm

solution

we take here equation of length that is

L = $n \times \frac{1}{4} \lambda$ ...............1

total length will be here

$L = \frac{\lambda}{2} + \frac{\lambda}{4}\\$

$L = \frac{3 \lambda }{4}$

so $\lambda$ will be

$\lambda = \frac{4L}{3}\\\lambda = \frac{4\times 30}{3}$

$\lambda$ = 40 cm

5 0

3 years ago

27.0 mL to the nearest milliliter

Mila [183]

Answer:

27.0 milliliters is the nearest mililiter so 27.0 is the answer

Explanation:

8 0

3 years ago

A caris initially at rest starts moving with a constant acceleration of 0.5 m/s2 and travels a distance of 5 m. Find

EleoNora [17]

Answer:

(I)

${ \bf{ {v}^{2} = {u}^{2} - 2as }} \\ {v}^{2} = {0}^{2} - (2 \times 0.5 \times 5) \\ {v}^{2} = 5 \\ { \tt{final \: velocity = 2.24 \: {ms}^{ - 1} }}$

(ii)

${ \bf{v = u + at}} \\ 2.24 = 0 + (0.5t) \\ { \tt{time = 4.48 \: seconds}}$

8 0

3 years ago

In a charging process, 4 × 1013 electrons are removed from one small metal sphere and placed on a second identical sphere. Initi

Alina [70]

Answer:

The distance between the two spheres is 914.41 X 10³ m

Explanation:

Given;

4 X 10¹³ electrons, and its equivalent in coulomb's is calculated as follows;

1 e = 1.602 X 10⁻¹⁹ C

4 X 10¹³ e = 4 X 10¹³ X 1.602 X 10⁻¹⁹ C = 6.408 X 10⁻⁶ C

V = Ed

where;

V is the electrical potential energy between two spheres, J

E is the electric field potential between the two spheres N/C

d is the distance between two charged bodies, m

$V = \frac{K*q}{d^2}*d = \frac{K*q}{d}$

$d = \frac{K*q}{V}$

where;

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

d = (8.99 X 10⁹ X 6.408 X 10⁻⁶)/0.063

d = 914.41 X 10³ m

Therefore, the distance between the two spheres is 914.41 X 10³ m

3 0

3 years ago

A solution containing 10.0 g of an unknown liquid and 90.0 g water has a freezing point of -3.33 °C. Given Kf = 1.86 °C/m for wa

Fantom [35]

Answer:

62.06 g/mol

Explanation:

We are given that a solution containing 10 g of an unknown liquid and 90 g

Given mass of solute = $W_B$ =10 g

Given mass of solvent= $W_A$ =90 g

$k_f=1.86^{\circ}C/m$

Freezing point of solution =-3.33 $^{\circ}$ C

Freezing point of solvent = $0^{\circ}$ C

Change in freezing point =Depression in freezing point

=Freezing point of solvent - freezing point of solution=0+3.33= $3.33^{\circ}$

$\Delta T_f=\frac{W_B\times K_f\times 1000}{W_A\times M_B}$

$M_B=\frac{10\times 1.86\times 1000}{3.33\times 90}$

$M_B=62.06 g/mol$

Hence, molar mass of unknown liquid is 62.06g/mol.

6 0

3 years ago

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