Help please !! I need help with this!

?during the first few days of a fast, what energy source provides about 90% of the glucose needed to fuel the body?

Kobotan [32]

The energy source provides about 90% of the glucose needed to fuel the body in the first few days of the fast is protein. protein is a string of amino acids that peforms a variety of functions to the body. These are found in foods like eggs, cheese, chicken breasts, meat and many more/.

8 0

3 years ago

An electron is moving horizontally east in an electric field that points vertically downward. The electric force on the electron

saveliy_v [14]

Answer:

d. )directed upward.

Explanation:

As the electron has a negative charge, when under the influence of an electric field, is subject to an electric force, which direction is the opposite to the direction of the electric field.

This is because the electric field has the same direction that the force on a positive test charge at the same point.

As the electric field points vertically downward, the electric force on the electron (a negative charge) points vertically upward.

So, the statement d. is the one that results to be true.

7 0

3 years ago

A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.

IrinaVladis [17]

Answer:

Approximately $722\; \rm m\cdot s^{-1}$ .

Explanation:

The average speed of a vehicle is calculated as:

$\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}$ .

In this question, the total distance is $2 \times 560\; \rm km = 1120\; \rm km$ .

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

$560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m$ .

$1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m$ .

Time required for the first part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s$ .

Time required for the second part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s$ .

The time required for the entire trip would be approximately $930 + 620 = 1550\; \rm s$ .

Calculate the average speed of this plane:

$\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}$ .

6 0

3 years ago

The Michelson-Morley experiment a) confirmed that time dilation occurs. b) proved that length contraction occurs. c) verified th

hram777 [196]

Answer:

e) indicated that the speed of light is the same in all inertial reference frames.

Explanation:

In 18th century, many scientists believed that the light just like air and water needs a medium to travel. They called this medium aether. They believed that even the space is not empty and filled with aether.

Michelson and Morley tried to prove the presence and speed of this aether through an interference experiment in 1887. They made an interferometer in which light was emitted at various angles with respect to the supposed aether. Both along the flow and against the flow to see the difference in the speed of light. But they did not find no major difference and thus it became the first proof to disprove the theory of aether.

It thus proved that the speed of light remains same in all inertial frames.

Also, it became a base for the special theory of relativity by Einstein.

5 0

3 years ago

Two identical metal balls a and b are mounted on insulating rods. Ball a has a charge of +q/2 and ball b is initially uncharged.

oksano4ka [1.4K]

Answer:

Help me please?

Explanation:

Did you get the answer? I believe it’s either C. +q or D. 0

6 0

3 years ago

Read 2 more answers

Help please !!I need help with this!