1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic

when a stationary rugby ball is kicked, it is contact with a player's about for 0.05 s. during this short time, the ball acceler

Ad libitum [116K]

Answer:

30 m/s

Explanation:

Applying,

v = u+at................ Equation 1

Where v = final speed of the ball, u = initial speed of the ball, a = acceleration, t = time.

From the question,

Given: u = 0 m/s (stationary), a = 600 m/s², t = 0.05 s

Substitute these values into equation 5

v = 0+(600×0.05)

v = 30 m/s

Hence the speed at which the ball leaves the player's boot is 30 m/s

3 0

3 years ago

What is conflict resolution?

enot [183]

Conflict resolution is conceptualized as the methods and processes involved in facilitating the peaceful ending of conflict and retribution.

7 0

3 years ago

A mule is harnessed to a sled having a mass of 246 kg, including supplies. The mule must exert a force exceeding 1210 N at an an

zimovet [89]

(a) 1800 N

The equation of the forces along the vertical direction is:

$F sin \theta + N - mg = 0$

where

$F sin \theta$ is the component of the applied force along the vertical direction

N is the normal force on the sled

mg is the weight of the sled

Substituting:

F = 1210 N

m = 246 kg

$\theta = 30.3^{\circ}$

We find N:

$N=mg-F sin \theta = (246)(9.8)-(1210)(sin 30.3^{\circ})=1800 N$

(b) 0.580

The equation of the forces along the horizontal direction is:

$F cos \theta - \mu_s N = 0$

where

$F cos \theta$ is the horizontal component of the push applied by the mule

$\mu_s N$ is the static frictional force

Substituting:

F = 1210 N

N = 1800 N

$\theta = 30.3^{\circ}$

We find $\mu_s$ , the coefficient of static friction:

$\mu_s = \frac{F cos \theta}{N}=\frac{(1210)(cos 30.3^{\circ})}{1800}=0.580$

(c) 522 N

In this case, the force exerted by the mule is

$F= 6.05 \cdot 10^2 N = 605 N$

So now the equation of the forces along the horizontal direction can be written as

$F cos \theta - F_f = 0$

where

$\theta=30.3^{\circ}$

and $F_f$ is the new frictional force, which is different from part (b) (because the value of the force of friction ranges from zero to the maximum value $\mu N$ , depending on how much force is applied in the opposite direction)

Solving the equation,

$F_f = F cos \theta = (605)(cos 30.3^{\circ})=522 N$

6 0

3 years ago

The electric potential inside a charged spherical conductor of radius R is is given by V = keQ/R, and the potential outside is g

Bumek [7]

Answer:

For outer points of shell

$E = \frac{k_eQ}{r^2}$

Now for inner point of shell

$E = 0$

Explanation:

As we know that out side the shell electric potential is given as

$V = \frac{K_e Q}{r}$

inside the shell the electric potential is given as

$V = \frac{K_e Q}{R}$

now we know the relation between electric potential and electric field as

$E = - \frac{dV}{dr}$

so we can say for outer points of the shell

$E = -\frac{dV}{dr}$

$E = - \frac{d}{dr}(\frac{K_eQ}{r})$

$E = \frac{k_eQ}{r^2}$

Now for inner point again we can use the same

$E = - \frac{dV}{dr}$

$E = - \frac{d}{dr}(\frac{K_eQ}{R})$

$E = 0$

6 0

3 years ago

A skydiver falls toward the ground at a constant velocity. Which statement best applies Newton’s laws of motion to explain the s

Tema [17]

Answer:

The answer is (a.) An upward force balances the downward force gravity on the skydiver

The skydiver is falling at a constant velocity because the upward force is balancing the downward force of gravity. According to Newton, the opposite force balance each other. This is stated in Newton's second law of motion.

7 0

3 years ago

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1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic​

1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic