Select the correct answer. What is tan for the given triangle? 0.60 1.67 1.33 5.00 0.20

What happens at the end of most cold currents?

Akimi4 [234]

Answer:

Ocean currents act as conveyer belts of warm and cold water, sending heat toward the polar regions and helping tropical areas cool off, thus influencing both weather and climate.

Explanation:

5 0

2 years ago

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An empty glass beaker has a mass of 103 g. When filled with water, it has a total mass of 361g.

aivan3 [116]

Answer:

0.96 gcm¯³

Explanation:

From the question given above, the following data were obtained:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of beaker + oil = 351 g

Density of water = 1 gcm¯³

Density of cooking oil =?

Next, we shall determine the mass of water. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of water =?

Mass of water = (Mass of beaker + water) – (Mass of empty beaker)

Mass of water = 361 – 103

Mass of water = 258 g

Next, we shall determine the volume of the beaker. This can be obtained by calculating the volume of water in the beaker.

Density of water = 1 gcm¯³

Mass of water = 258 g

Volume of water =?

Density = mass /volume

1 = 258 / volume

Cross multiply

1 × volume = 258

Volume of water = 258 cm³

Thus the volume of the beaker is 258 cm³.

Next, we shall determine the mass of the cooking oil. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + oil = 351 g

Mass of cooking oil =?

Mass of cooking oil = (Mass of beaker + oil) – (Mass of empty beaker)

Mass of cooking oil = 351 – 103

Mass of cooking oil = 248 g

Finally, we shall determine the density of the cooking oil. This can be obtained as follow:

Mass of cooking oil = 248 g

Volume of the beaker = 258 cm³

Density of cooking oil =?

Density = mass / volume

Density = 248 / 258

Density of cooking oil = 0.96 gcm¯³

7 0

3 years ago

An archer shoots an arrow with a mass of 45.0 grams from bow pulled

Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

3 0

2 years ago

Q 1: Calculate the pressure.

Oksana_A [137]

Answer:

1: 300pa

2: 10cm2

3: 430pa

4: a: 1.6667pa

b:2.5m2

c:20pa

Explanation:

7 0

3 years ago

Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter

Aleonysh [2.5K]

We can solve the problem by using the first law of thermodynamics:

$\Delta U= Q-W$

where

$\Delta U$ is the variation of internal energy of the system

Q is the heat added to the system

W is the work done by the system

In this problem, the variation of internal energy of the system is

$\Delta U=U_f-U_i=134 J-87 J=47 J$

While the heat added to the system is

$Q=500 J$

therefore, the work done by the system is

$W=Q-\Delta U=500 J-47 J=453 J$

5 0

3 years ago

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