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2 years ago

Acceleration of a particle moving along x axis is given

Physics

1 answer:

amm18122 years ago

7 0

B. The velocity of the particle with respect to the acceleration is √x + 4.

<h3>What is velocity?</h3>

Velocity is the rate of change of displacement with time.

The velocity of the particle with respect to the acceleration of the particle is calculated as follows;

vf = v₀ + at

where;

vf is the final velocity
v₀ is the initial velocity
a is acceleration, assuming a = √x + 2
t is time

vf(x) = 2 + (√x + 2)

vf(x) = √x + 2 + 2

vf(x) = √x + 4

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A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor

Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force $F_{1}=6\ N$

Second force $F_{2}=10\ N$

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

$\tau=Force\times lever\ arm$

So, Here,

$\tau_{2}=5 \tau_{1}$

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

$F_{2}\times d_{2}=5\times F_{1}\times r_{1}$

$r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}$

Where, $F_{1}$ =First force

$F_{1}$ =First force

$F_{2}$ =Second force

$r_{1}$ = distance

Put the value into the formula

$r_{2}=\dfrac{5\times6\times0.2}{10}$

$r_{2}=0.6\ m$

Hence, The moment arm is 0.6 m

6 0

3 years ago

1-A train travels 100 km to reach town A in one hour and 15 min. The train stops at station A for 45 minutes. Then it travels 15

kirill [66]

Answer 1) : 62.5 km/hour is the average velocity of the train.

2) The final velocity of the car at the end of 75 m is 14.69 m/s

Explanation:

1) Displacement of the train = 100 km + 150 km = 250 km

Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours

Average velocity= $\frac{Displacement}{time}=\frac{250 km}{4 hour}=62.5 km/h$

62.5 km/hour is the average velocity of the train.

2) The acceleration of the car, a= 1.2 $m/s^2$

Distance covered by the car,s = 75 m

Initial velocity of the car , $v_i$ = 6 m/s

Final velocity of thre car , $v_f$ =?

Using third equation of motion:

$v_{f}^2=v_{i}^2+2as=(6 m/s)^2+2\times 1.2 m/s\times 75 m=216 m^2/s^2$

$v_{f}=14.69 m/s$

The final velocity of the car at the end of 75 m is 14.69 m/s

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3 years ago

Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way

VladimirAG [237]

Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E

The electric field between the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

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3 years ago

2. For a rotating rigid body, which of the following statements is NOT correct?

AfilCa [17]

Answer:

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3 years ago

if a ball is thrown straight up into the air with an initial velocity of 8080 ft/s, its height in feet after tt second is given

yanalaym [24]

The average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds = 63.84 ft/s

(ii) 0.001 seconds = 63.984 ft/s

Given that a ball is thrown with an initial velocity = 80 ft/s

Let 'y' be the height in feet after 't' seconds.

Given, $y=80t-16t^2$ gives the height in 't' seconds.

Average velocity = Rate of change of distance

= Change in distance/Change in time.

The initial time can be taken as 0 s.

When t =1 s, y = 80 - 16 = 64 ft

(1) t = 0.01 s

y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft

Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s

(2) t = 0.001 s

y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft

Average velocity = (64 - 0.079984) / (1 -0.001) = 63.984 ft/s

The question is incomplete. Find out the complete question below:

If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by $y=80t-16t^2$ .Find the average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds

(ii) 0.001 seconds

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2 years ago