Find the force on an object which has a mass of 20 kg and an acceleration of 10 m/s2.

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and

garri49 [273]

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

(c) vf = 7.17 m/s

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

$2gh = v_f^2 - v_i^2$

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

$(2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\$

h = 0.82 m

Now, for the time in air during upward motion we use first equation of motion:

$v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s$

(c)

Now we will consider the downward motion and use the third equation of motion:

$2gh = v_f^2-v_i^2$

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

$2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\$

vf = 7.17 m/s

Now, for the time in air during downward motion we use the first equation of motion:

$v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s$

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

t = 1.14 s

7 0

2 years ago

If a force of 3000 N is applied to a large rock, but the rock does not move, how much work is done on the rock?

kkurt [141]

Answer:

No work was done.

W = 0

Explanation:

Work is said to be done whenever a force of one newton moves a body of one kilogram through a distance of one meter. Meaning the applied force has to move the body from a point of rest through certain distance.

Work = force × distance

So, in the case of this question, we only have the force been applied, but no distance was covered. Hence, no work was done.

W = 3000× 0 meter

W = 0

8 0

2 years ago

A plank rests on top of the axles of two identical wheels. Each wheel's outer radius is 0.25 meters and each wheel's axle has ra

kaheart [24]

Answer:

The plank moves 0.2m from it's original position

Explanation:

we can do this question from the constraints that ,

the wheel and the axle have the same angular speed or velocity
the speed of the plank is equal to the speed of the axle at the topmost point .

thus ,

since the wheel is pure rolling or not slipping,

⇒ $v=wr$

where

$v$ - speed of the wheel

$w$ - angular speed of the wheel

$r$ - radius of the wheel

since the wheel traverses 1 m let's say in time ' $t$ ' ,

$v_{w}=\frac{distance}{time} =\frac{1}{t}$

∴

⇒ $w=\frac{v_{w}}{r} = \frac{1}{t*0.25}$

the speed at the topmost point of the axle is :

⇒ $v_{a}=w*r\\v_{a}=\frac{1}{t*0.25} *0.05\\v_{a}=\frac{1}{5t}$

this is the speed of the plank too.

thus the distance covered by plank in time ' $t$ ' is ,

⇒ $d=v_{a}*t\\d=\frac{1}{5t} *t\\d=\frac{1}{5} = 0.2m$

8 0

3 years ago

In an electrical circuit 200mC of charge flows through a 50kohm resistor in 90 seconds. What is the power

ss7ja [257]

Answer: $0.2464\ W$

Explanation:

Given

Charge $Q=200\ mC$

Resistance $R=50\ k\Omega$

Time $t=90\ s$

Charge is the amount of current flown in a particular time

$Q=It\\\Rightarrow 200\times 10^{-3}=I\times 90\\\\\Rightarrow I=\dfrac{200\times 10^{-3}}{90}\\\\\Rightarrow I=2.22\times 10^{-3}\ A$

Power is the given by $P=I^2R$

Insert values

$\Rightarrow P=\left(2.22\times 10^{-3}\right)^2\times 50\times 10^3\\\\\Rightarrow P=246.42\times 10^{-3}\ W\\\\\Rightarrow P=0.2464\ W$

The power is in the circuit is $0.2464\ W$ .

8 0

2 years ago

sabendo que uma onda sonora tem um periodo de 0,002s e um comprimento de onda de 0,5m determine sua velocidade nesse meio

Artemon [7]

Answer: Sorry i need more Details. I want to help but I dont speak spanish.

Explanation: Maybe come back later with a english question. So sorry. I hope ypu have a good Day.

3 0

2 years ago