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3 years ago

Mass of object is 50g moves in a circular path of radius 10cm find work done

Physics

1 answer:

topjm [15]3 years ago

3 0

Answer:

Work done = 0.3142 Nm

Explanation:

Mass of Object is 50 g

Circular path of radius is 10 cm ⇒ 0.1 m

Work done = Force × Distance = ?

*Distance moved (circular path) ⇒ Circumference of the circular path

2πr = 2 × 3.142 × 0.1 ⇒ 0.6284 m

*Force that is enough to move a 50 g must be equal or more than its weight.

therefore convert 50 grams to newton = 0.5 N

Recall that; work done is force times distance

∴ 0.5 N × 0.6284 m

Work done = 0.3142 Nm

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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

3 years ago

Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on

qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

Δx=760 m
V₀=87 m/s
t=13.6 s
a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

$760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}$

$760=(1183.2)(cos\theta)$

$cos\theta=\frac{760}{1183.2}$

$\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}$

3 0

3 years ago

You walk to the north, then turn 90° to your left and walk another How far are you from where you originally started?

rodikova [14]

Whatever distance north and then west you walked, you are then

(1.41 x that distance)

northwest of where you started.

3 0

3 years ago

Explain the process of synaptic transmission, beginning with the neurotransmitters at the axon terminal of the presynaptic cell

nydimaria [60]

The synapse is actually the link between 2 neurons. Now when an action potential contacts the synaptic knob of a neuron, the voltage-gate calcium channels are unlocked, resulting in an influx of positively charged calcium ions into the cell. This makes the vesicles containing neurotransmitters, for example acetylcholine, to travel towards the pre-synaptic membrane. When the vesicle arrives at the membrane, the contents are released into the synaptic cleft by exocytosis. Neurotransmitters disperse across the space, down to its concentration gradient, up until it reaches the post-synaptic membrane, where it connects to the correct neuroreceptors. Connecting to the neuroreceptors results in depolarisation in the post-syanaptic neuron as voltage-gated sodium channels are also opened, and the positively charged sodium ions travel into the cell. When adequate neurotransmitters bind to neuroreceptors, the post-synaptic membrane overcame the threshold level of depolarisation and an action potential is made and the impulse is transmitted.

8 0

3 years ago

Read 2 more answers

Devise an exponential decay function that fits the given data, then answer the accompanying questions. Be sure to identify the

7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

$P=P_0e^{-kh}$

where,

$P_0$ =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

$P=\dfrac{1}{3}P_0$

$\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}$

Now

$P=\dfrac{1}{2}P_0$

$ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft$

The altitude will be 22145.27733 ft

$P=0.02P_0$

$0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft$

The elevation is 124984.76055 ft

6 0

3 years ago