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3 years ago

Any help please, physics is not my strong suit?

Physics

2 answers:

frez [133]3 years ago

5 0

Answer:

2000mg = 2g

5L = 5000mL

16cm = 160mm

Mashcka [7]3 years ago

5 0

1) kg

2) m

3) g

4) mL

5) mm

6) L

7) km

8) cm

9) mg

10) 2g

11) 104000m

12) 4.8m

13) 5600g

14) 8cm

15) 5000mL

16) 0.198kg

17) 0.075L

18) 0.15m

19) 560cm

20) 16mm

21) 25km

22) 6500mg

23) 66mm

24) 0.12g

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Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 19.0 ∘ west of north,

valina [46]

Answer:

$=2.99\times10^9J$

Explanation:

According to the question

net force F = 2.20×10^6 N

displacement $S = 0.72\times10^3m$

from figure , the horizontal forces are same in magnitude and opposite direction.

so , neglect these two forces.

we can take only vertical components of the force.

total force F' = F cos 19° + F cos 19°

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therefore , total work is

W = F'S

= (2F cos19)×S

$= (2)(2.20\times10^6 N)cos19° (0.720\times10^3 m)$

$=2.99\times10^9J$

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3 years ago

You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.How muc

ahrayia [7]

Work is equal to the force applied times the displacement. Since you pull the wagon at constant speed this means that there is no acceleration on the wagon as it does not change speed. F=ma. Since a=0, F=0. Therefore no work has been done in this situation

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3 years ago

when charge 2 is 3.0 m away from charge 1, the strength of the electric force on charge 2 by charge 1 is 0.80 n. if instead, cha

Nikolay [14]

The strength of the electrostatic force is inversely proportional to the square of the distance

between the charges. If the distance is doubled, the force is 1/4.

The new force is 0.80/4 = 0.20 N.

<h3>What is electrostatic force? </h3>

Electrostatic force is the attractive or repulsive force that exists between two charged particles. Also known as Coulomb interaction or Coulomb force. For example, the electrostatic forces between the protons and electrons of an atom are responsible for the stability of the atom.

The force acting along a line joining two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

$F=K |\frac{q_{1} q_{2}}{r^{2} } |$

In the above formula, k is arbitrary and can be chosen to be any positive value. Since k is a constant, I chose to give the value of k as follows:

Therefore, with q₁ and q₂ values of 1 and r = 1 (two charges with 1 Coulomb charge each at a distance of 1 m), we get F = $9 \times 10^9$ N. In the above equation, ε₀ is given as the permittivity of free space and its value in SI units is $8.854\times10^{-12} C^{2} N^{-1} m^{-2}$ .

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1 year ago

A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef

bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

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a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

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3 years ago

A rolling ball has an initial velocity of 1.6 meters per second. if the ball has a constant acceleration of 0.33 meters per seco

lukranit [14]

We have:

Initial velocity (u) = 1.6 m/s
Constant acceleration (a) = 0.33 m/s²
Time (t) = 3.6 sec

There are five constant acceleration equations that would help us to find the velocity:

$v=u+at$
$s=ut+ \frac{1}{2}at^2$
$s= \frac{1}{2}(u+v)t$
$v^2=u^2+2as$
$s=vt- \frac{1}{2}at^2$

Since we have $u, a, t$ and we want $v$
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$v=1.6+(0.33)(3.6)$
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7 0

2 years ago