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AfilCa [17]
2 years ago
6

2. An overseas jet requires 6 hours to fly 9700 km. What is the jet's speed?

Physics
1 answer:
Debora [2.8K]2 years ago
4 0

Answer:

Most commercial aircraft typically fly at around 460-575 mph, or 740-930 km/h, according to Flight Deck Friend. But private jet speed can vary depending on a variety of factors, such as the weight onboard and the weather conditions.

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How much force is required to pull a spring 3.0 cm from
avanturin [10]

Answer:

I know that T= kx where T is the tension which equaka the force og gravity = mg = 1.37 * 10 = 13.7 x is the elongation of the spring so the length after dangling the object minus the original length.

I hope it helps

plz let me know if it is wrong or right.

4 0
4 years ago
Julie wants to know if adding calcium fertilizer to tomato plants will help prevent a disease of tomato plants called blossom-en
Ksivusya [100]
It's not true.

The crucial principle for a scientific experiment is to keep only ONE variable at a time.

In this case, the variable of this experiment is actually the tomato is in sunny part or in shady part, instead of whether applying Ca fertilizer. 
3 0
3 years ago
A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the si
solmaris [256]

Answer:

4.384 * 10^13

Explanation:

Given the expression :

[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]

Applying the laws of indices

[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]

13.2733 * 10^19 ÷ 3.0276 * 10^6

(13.2733 / 3.0276) * 10^(19 - 6)

4.3840996 * 10^13

= 4.384 * 10^13

6 0
3 years ago
1. A woman driving in a car at 21 m/s pushes down on the gas pedal. 6 seconds later, she is moving at 36 m/s. What was her avera
Furkat [3]

a=Vf-Vi/t=36-21/6=15/6=2.5m/s^2

5 0
3 years ago
An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the
Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m

(a) Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2

(b) Magnification, m=\dfrac{h'}{h}

h' is image height

-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m

Hence, this is the required solution.

4 0
3 years ago
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