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3 years ago

Is grade 10 harder or 9? I passed 9 grade thise year by 89,36 my mark..im scared of high school plis help and some advice ?

1 answer:

4 0

Listen in all of your classes and write down notes on EVERYTHING even if it may not seem important it will be useful in the future! :)

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A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?

olya-2409 [2.1K]

Answer:

First of all the formula is F= uR,( force= static friction× reaction)

mass= 5+25=30

F= 50

R= mg(30×10)=300

u= ?

F=UR

u= F/R

u= 50/300=0.17N

3 0

3 years ago

Help with 3 please! I will give brainliest!

rusak2 [61]

The answer is point b because vertical velocity is zero at the maximum height

8 0

3 years ago

Identical spheres are dropped from a height of 100 m above the surfaces of Planet X and Planet Y. The speed of the spheres as a

Anna007 [38]

Answer:B

Explanation:

8 0

3 years ago

Your boat departs from the bank of a river that has a swift current parallel to its banks. If you want to cross this river in th

REY [17]

Answer:

Explanation:

Given that your boat departs from the bank of a river that has a swift current parallel to its banks. If you want to cross this river in the shortest amount of time, you should direct your boat: so that it drifts with the current.

If the boat moves perpendicular to the current, the current flow will be the resistance to the movement of the boat. So, it's better for the boat to drifts perpendicularly with the current.

The best answer is therefore option D.

8 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

2 years ago