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timurjin [86]
2 years ago
9

A cat runs 80 meters to the left in 16 seconds, then 50 meters to the right in 10 seconds, then 70 meters to the left in 14 seco

nds, then 40 m to the right in 8 seconds.
a. What was the distance travelled by the cat?

b. What was the displacement of the cat?

c. What was the average speed of the cat?

d. What was the average velocity of the cat?
Physics
1 answer:
11Alexandr11 [23.1K]2 years ago
8 0
<h3>Answer:a) Distance= 240m. b) Displacement= 90m. c) Average Speed= 5ms^{-1}. d) Average Velocity= 1.875 ms^{-1}</h3>

Explanation:

a) Distance ➤ (80+50+70+40) m. ➤ Distance= 240m.

b) Displacement ➤ (50+40)m. ➤ Displacement= 90m

c) Average Speed ➤ Total distance travelled/total time taken

                              ➤ 240/48

                              ➤ 5 ms^{-1}

d) Average Velocity ➤ Total displacement/total time taken

                                 ➤ 90/48

                                 ➤ 1.875 ms^{-1}

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Ierofanga [76]
It looks blue as it is only reflecting blue light
6 0
2 years ago
A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu
romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand  = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 =  0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0
2 years ago
Use the equation for magnetic force on a moving charge to derive the equation for magnetic force on a current carrying wire. Sho
max2010maxim [7]

Answer:

The formula comes from Lorentz force law which includes both the electric and magnetic field. If the electric field is zero, the force law for just the magnetic field is <u>F=q(ν×B</u>) . Here, F  is force and is a vector because the force acts in a direction.  q  is the charge of the particle.  v  is velocity and is a vector because the particle is moving in some direction.  B is the magnetic flux density.

We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity vd.  Since the magnitude of B is constant at every line element of the loop (circle) and it dot product with the line element is B dl everywhere, therefore

                                                  ∮B dl=μ0 I

                                                  B ∮dl=μ0 I

                                                  B 2πr=μ0 I

                                                   B=μ02πr Id=μ0/4π I dl×rr3

Since, r can be written as r=(rcosθ,rsinθ,z) and dl as dl=(dl,0,0) And now, if we take the cross product we would get

                                               dl×r=−z dlj^+rsinθk^

and therefore the magnitude of dB is equal to

dB=μ0/4π I |dl×r|/r3=μ0/4π I z2+r2sin2θ−−−−−−−−−−√dl/r3

Thus, magnetic field is depending on r,θ,z.

Learn more about Force here-

brainly.com/question/2855467

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7 0
1 year ago
A circular sign has a diameter of 40 cm and is subjected to normal winds up to 150 km/h at 10°C and 100 kPa. Determine the drag
Marat540 [252]

Answer:

147.7 N

221.55 Nm

Explanation:

P = Pressure = 100000 Pa

R_s = Mass-specific gas constant = 287.015 J/kg k

T = Temperature = 10+273 = 283 K

C = Drag coefficient = 1.1

A = Area

r = Radius = 0.2 m

v = Speed of wind = \frac{150}{3.6}\ m/s

L = Length of pole

Density

\rho=\frac{P}{R_sT}\\\Rightarrow \rho=\frac{100000}{287.058\times 283}\\\Rightarrow \rho=1.2309\ kg/m^3

Drag force

F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2309\times 1.1\times \left(\pi \times 0.2^2\right)\times \left(\frac{150}{3.6}\right)^2\\\Rightarrow F=147.7\ N

Force on the circular sign is 147.7 N

M=F\times L\\\Rightarrow M=147.7\times 1.5\\\Rightarrow M=221.55\ Nm

Bending moment at the bottom of the pole is 221.55 Nm

6 0
3 years ago
An ac series circuit has an impedance of 60 Ohm and
Iteru [2.4K]

Answer:

Power factor of the AC series circuit is cos\phi=0.5

Explanation:

It is given that,

Impedance of the AC series circuit, Z = 60 ohms

Resistance of the AC series circuit, R = 30 ohms

We need to find the power factor of the circuit. It is given by :

cos\phi=\dfrac{R}{Z}

cos\phi=\dfrac{30}{60}

cos\phi=\dfrac{1}{2}

cos\phi=0.5

So, the power factor of the ac series circuit is cos\phi=0.5. Hence, this is the required solution.

6 0
2 years ago
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