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3 years ago

9. Calculate the distance (in km) that Charlie runs if he maintains the average

Physics

1 answer:

Karo-lina-s [1.5K]3 years ago

5 0

Correct Question:

Calculate the distance (in km) charlie runs if he maintains an average speed of 8 km/hr for 1 hour

Answer:

The total distance covered by Charlie is 8 km in 1 hour.

Explanation:

The average velocity as given in the question is,

v = 8 km/hr

Total time taken,

$$t=1 hour$

As we know the formula to evaluate the total distance d when the average velocity and time is given;

$v=\frac{d}{t}$

$d=v \times t$

$d=8 \times 1$

$d=8 k m$

Hence, the total distance covered by Charlie in 1 hour will be 8 km.

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two masses are kept 2 metre apart there is gravitational force of 2 Newton what is the gravitational force when they are kept at

sukhopar [10]

Answer: 0.5N

Explanation:

Gravitational force is calculated using the formula :

F = Gm1m2/r^2

Where G is the gravitational constant (6.67 × 10^-11)

At a distance 'r' of 2metres apart:

Mass of objects are m1 and m2

Gravitational force 'F1' = 2N

Inputting values into the formula :

2 = Gm1m2 / 2^2 - - - - - (1)

At a distance 'r' of 4meters apart:

Mass of objects are m1 and m2

Gravitational force 'F2' = y

Inputting values

F2 = Gm1m2 / 4^2 - - - - - (2)

Dividing equations 1 and 2

2 = Gm1m2 / 2^2 ÷ F2 = Gm1m2 / 4^2

2 / F2 = (Gm1m2 / 4) / (Gm1m2 / 16)

2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)

2/F2 = 16 / 4

Cross multiply

2 × 4 = 16 × F2

8 = 16F2

F2 = 8/16

F2 = 0.5N

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3 years ago

Sug . To repeat the experiment what is it?

statuscvo [17]

Answer:

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Explanation:

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3 years ago

You and a friend each carry identical boxes from the first floor of a building to a room located on the second floor, farther do

melamori03 [73]

Answer:

Work done in both the cases will be same

Explanation:

As we know that the work done against gravity is given as

$W = F_g .d$

here we know that gravitational force is a conservative force and the work done against gravitational force is independent of the path

So here the work done by person to move the object between two different heights will be independent of the path they choose

So for the first person and second person will be same in both the cases because the height through which the boxes are transferred will be same in both the cases

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3 years ago

A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (

Anika [276]

Answer:

$a=9.8 rad/s^{2}$

Explanation:

Torque, $\tau$ is given by

$\tau=Fr$ where F is force and r is perpendicular distance

$R=0.5Lcos\theta$ where $\theta$ is the angle of inclination

Torque, $\tau$ can also be found by

$\tau=Ia$ where I is moment of inertia and a is angular acceleration

Therefore, Fr=Ia and F=mg where m is mass and g is acceleration due to gravity

Making a the subject, $a=\frac {Fr}{I}=\frac {mgr}{I}$ and already I is given as

$I=\frac {mL^{2}}{3} and r is 0.5Lcos\theta$ hence

$a=\frac {0.5mgLcos\theta}{1/3 mL^{2}}$

$a=\frac {3gcos\theta}{2L}$

Taking g as 9.81, $\theta$ is given as 37 and L is 1.2

$a=\frac {3*9.81cos37}{2*1.2}=9.7932679419$

$a=9.8 rad/s^{2}$

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3 years ago

The 10-lb block has a speed of 4 ft/s when the force of f=(8t2)f=(8t2) lb is applied. determine the velocity of the block when t

KatRina [158]

The velocity of the block when t == 2 s is 60.7 ft./sec.

Equations of Motion.

Here the friction is $F_f = \mu_k N$ = 0.2 N

$+ \uparrow \sum F_y = ma_y; \quad N – 10 = \frac { 10 } { 32.2 }(0) \quad N = 10 lb \\ \begin{aligned} \underrightarrow{ + } \sum F_x = ma_x; \quad 8t^2 – 0.2(10 &) = \frac { 10 } { 32.2 }a \\ & a = 3.22(8t^2 – 2) ft/s^2 \end{aligned}$

Kinematics.

The velocity of the block as a function of t can be determined by

integrating dv = adt using the initial condition v = 4 ft./s at t = 0.

$\int_{ 4 ft/s }^{ v } dv = \int_0^t 3.22(8t^2 – 2)dt \\ \begin{aligned} v – &4 = 3.22 (\frac 8 3 t^3 – 2t) \\ & v = \{8.5867t^3 – 6.44t + 4 \} ft/s \end{aligned}$

The displacement as a function of t can be determined by integrating

ds = vdt using

the initial condition s = 0 at t = 0

$\int_0^s ds = \int_0^t (8.5867t^3 – 6.44t + 4)dt \\ s = \{2.1467t^4 – 3.22t^2 + 4t \} ft$

at t = 2 sec

s = 30 ft.

Thus, at s = 30 ft.,

$\begin{aligned} v &= 8.5867(2.0089^3) – 6.44(2.0089) + 4 \\ &= 60.67 ft/s \\ &= 60.7 ft/s \end{aligned}$

Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move.

Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of mathematics. A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system.

Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined. The study of how forces act on bodies falls within kinetics, not kinematics. For further details, see analytical dynamics.

Learn more about kinematics here : brainly.com/question/24486060

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1 year ago