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3 years ago

11

Please answer correctly to be marked as brainly.

1 answer:

lidiya [134]3 years ago

7 0

A- PRICE

B-QUANTITY

C-SUPPLY

D-DEMAND

E-EQUILIBRIUM POINT

Explanation:

It is the Supply Demand curve in Economics. It gives relationship between price and quantity

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Vector vector b has x, y, and z components of 4.00, 4.00, and 2.00 units, respectively. calculate the magnitude of vector

Sav [38]

Good morning.

We see that $\mathsf{\overset{\to}{b}} = \mathsf{(4.00, \ 4.00, \ 2.00)}$

The magnitude(norm, to be precise) can be calculated the following way:

$\star \ \boxed{\mathsf{\overset{\to}{a}=(x, y,z)\Rightarrow ||\overset{\to}{a}|| = \sqrt{x^2+y^2+z^2}}}$

Now the calculus is trivial:

$\mathsf{\|\overset{\to}{b} \| =\sqrt{4^2+4^2+2^2} =\sqrt{16+16+4}}\\ \\ \mathsf{\|\overset{\to}{b}\|=\sqrt{36}}\\ \\ \boxed{\mathsf{\|\overset{\to}{b}\| = 6.00 \ u}}$

7 0

3 years ago

An acorn falls from an oak tree. You note that it

Vaselesa [24]

Answer:9.8 m/s²

Explanation:

It was going at 9.8m/s² as this is the acceleration of an object due to gravity

when an object falls it accelerates at a consant and uniform speed which is 9.8m/s²

6 0

3 years ago

A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart

Musya8 [376]

1) $0.61 m/s^2$

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

$g=\frac{GM}{r^2}$ (1)

where

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

$r=4R$

where

$R=6.37\cdot 10^6 m$ is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

$g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2$

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

$g=0.61 m/s^2$ is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

$T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s$

4 0

3 years ago

When 525 J of heat is added to 7.6 g of metal at 26°C, the temperature increases to 80°C. What is the specific heat of the metal

Temka [501]

Ans the answer is 0.25J

Explanation

i guess

4 0

3 years ago

Suppose a radio signal (light) travels from Earth and through space at a speed of 3 × 10^8/ (this is the speed of light in vacuu

zlopas [31]

Answer:

Explanation:

we know that

s=vt here v is the speed and s is distance covered by the signals

given data

v=3*10^8

t=10 min we have to convert it into seconds

1 minute=60 seconds

so

10 minutes =10*60/1 =600 seconds

now putting the value of v and t we can find the value of s

s=vt

s=3*10^8*600

s=1.8*10^11m

i hope this will help you

8 0

3 years ago