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3 years ago

What causes the formation of ocean currents?

1 answer:

5 0

Ocean currents can be generated by wind, density differences in watermasses caused by temperature and salinity variations, gravity, and events such as earthquakes

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Gravity: The force due to gravity is F=mg where g=9.80 m/s². A. Find the force due to gravity on a 41.63-kg object. B. The force

olga_2 [115]

A. F=mg
= 41.63kg x 9.8m/s²
= 407.974
= 408N (3 significant figure)
B. F=mg
632N= y x 9.8m/s²
y= 632 ÷ 9.8
y = 64.48
y= 64.5kg (3 significant figure)

4 0

3 years ago

A pendulum swings back and forth 5 times in 10 seconds what is the period of the pendulum?

Ainat [17]

Period is T = 1/f. The frequency, f, is 5cycles/10s = 0.5. So the period T=1/0.2 = 2.

4 0

3 years ago

A current of 5 A passes through a variable resistor set to 15 Ω. Calculate the voltage

OLEGan [10]

Answer:

75 volt

Explanation:

Current (I) = 5 A

Resistance (R) = 15 Ω

Voltage (V) = ?

We know

R = V/I

15 = v / 5

v = 75 Volt

3 0

3 years ago

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b

Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy= $\frac{1}{2} kx^{2} +PE1$

Energy at the point where the block will stop consists of only gravitational potential energy= $PE2$

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒ $\frac{1}{2} kx^{2} +PE1=PE2$

⇒ $PE2-PE1=\frac{1}{2} kx^{2}$

Also $PE2-PE2=mgh$

where $m$ is the mass of block

$g$ is acceleration due to gravity=9.8 m/s

$h$ is the difference in height between two positions

⇒ $mgh=\frac{1}{2} kx^{2}$

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴ $2100*9.8*h=\frac{1}{2}*2200*0.11^{2}$

⇒ $20580*h=13.31$

⇒ $h=\frac{13.31}{20580}$

⇒h=0.0006467m= $6.5e-4$

7 0

3 years ago

A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone

Lemur [1.5K]

Answer:

Fy = 14.3 [N]

Explanation:

To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:

When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

$F = \sqrt{F_{x}^{2} +F_{y}^{2} }$

where:

F = 15 [N]

Fx = horizontal component = 4.5 [N]

Fy = vertical component [N]

$15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N]$

7 0

3 years ago