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strojnjashka [21]

3 years ago

8

15pts! brainliest to who answers!If Kyla picks up a grocery bag, using 10 N of force to lift it 1.5 m off the floor, how much wo

rk did Kyla do on the bag?

1 answer:

omeli [17]3 years ago

6 0

Explanation:

work = force x distance

w = 10 x 1.5 = 15Nm

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The equation below is for potassium oxide.

ValentinkaMS [17]

Answer: The ratio of atoms of potassium to ratio of atoms of oxygen is 4:2

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed, and remains conserved. The mass of products must be same as that of the reactants.

Thus the number of atoms of each element must be same on both sides of the equation so as to keep the mass same and thus balanced chemical equations are written.

K exists as atoms and oxygen exist as molecule which consists of 2 atoms. The ratio of number of atoms on both sides of the reaction are same and thus the ratio of atoms of potassium to ratio of atoms of oxygen is 4:2.

5 0

3 years ago

Read 2 more answers

Which is a scalar quantity displacement distance force acceleration

cestrela7 [59]

A scalar is a quantity that is fully described by a magnitude only. It is described by just a single number. Some examples of scalar quantities include speed, volume, mass, temperature, power, energy, and time. A vector is a quantity that has both a magnitude and a direction.

I hope this helps you.

5 0

3 years ago

Does anyone know this???

Gelneren [198K]

Answer:

C

Explanation:

8 0

3 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago

2. A portion of the body receives 0.15 mGy from radiation with a quality factor Q = 6 and 0.22 mGy from radiation with Q = 10. (

DiKsa [7]

Answer with Explanation:

We are given that

$D_1=0.15mGy$

$D_2=0.22mGy$

$Q_1=6$

$Q_2=10$

a.We have to find the total dose

Total dose= $D=D_1+D_2$

Using the formula then, we get

$D=0.15+0.22$

$D=0.37mGy$

b.We have to find the total dose equivalent

Total dose equivalent=H= $D_1Q_1+D_2Q_2$

Using the formula

$H=0.15(6)+0.22(10)$

H=3.1mSv

7 0

3 years ago