2 years ago

Same diagram... Which location shows SUMMER in the SOUTHERN hemisphere? *

Physics

2 answers:

Maksim231197 [3]2 years ago

8 0

Location one is the answer

LiRa [457]2 years ago

7 0

Location 1 is correct

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5. What is the amount of force required to accelerate a 20 kg object at a rate of 5 m/sz?

GenaCL600 [577]

Force required is 100 N

Given that;

Rate of acceleration = 5 m/s²

Mass of object = 20kg

Find:

Force required

Computation:

Force = Mass × Acceleration

Force required = Rate of acceleration × Mass of object

Force required = 20 × 5

Force required = 100 N

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What do we call the bright, sphere-shaped region of stars that occupies the central few thousand light-years of the milky way ga

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Answer: The galaxy's bulge

Explanation:

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1 year ago

The following items are connected in a circuit. in which one would you find a voltage gain?

timurjin [86]

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3 years ago

A food department is kept at â12Â°c by a refrigerator in an environment at 30Â°c. the total heat gain to the food department is

slava [35]

As per energy conservation in the reversible engine we can say

$Q_2 + W = Q_1$

here we know that

$Q_2 = 3300 kJ/h$

$Q_1 = 4800 kJ/h$

now from above equation

$3300 + W = 4800$

$W = 1500 kJ/h$

now we can convert it into kW

$W = 1500\times \frac{kJ}{3600s}$

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so above is the power input to the refrigerator

now to find COP we know that

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2 years ago

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ ...Newtons second law Fnet = ma

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ ...f is the friction which is zero here.

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0

2 years ago