تقطع اولا مسافة 8 km شمالا من البيت ثم تمشي شرقا حتى تكون ازاحتك من البيت 10km ما مقدار المسافة التي قطعتها شرقا

A dock worker applies a constant horizontal force of 81.0 N to a block of ice on a smooth horizontal floor. The frictional force

AnnZ [28]

Answer:

The correct answer is:

(a) 84.240 kg

(b) 24.038 m

Explanation:

The given values are:

Force,

F = 81.0 N

Distance,

S = 13.0 m

Time,

t = 5.20 s

As we know,

The acceleration of mass will be:

⇒ $a=\frac{2S}{t^2}$

On substituting the given values, we get

⇒ $=\frac{2\times 13.0}{(5.20)^2}$

⇒ $=\frac{26}{27.04}$

⇒ $=0.961538 \ m/s^2$

(a)

The mass of the block will be:

⇒ $m=\frac{F}{a}$

On substituting the given values, we get

⇒ $=\frac{81.0}{0.961538}$

⇒ $=84.240 \ kg$

(b)

The final velocity after a given time i.e.,

t = 5.00 s

⇒ $v=at$

On substituting the values, we get

⇒ $=0.961538\times 5.00$

⇒ $=4.8076 \ m/s$

In time, t = 5.00 s

The distance moved by the block will be:

⇒ $d=vt$

On putting the values, we get

⇒ $=4.8076\times 5.00$

⇒ $=24.038 \ m$

5 0

3 years ago

You stand on top a building 44 m tall with a water balloon. You drop the water balloon from rest. How fast is the balloon moving

BabaBlast [244]

Y₀ = initial position of the balloon at the top of the building = 44 m

Y = final position of the balloon at halfway down the building = 44/2 = 22 m

a = acceleration of the balloon = - 9.8 m/s²

v₀ = initial velocity of the balloon = 0 m/s

v = final velocity of the balloon = ?

using the kinematics equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 0² + 2 (- 9.8) (22 - 44)

v = 20.78 m/s

4 0

4 years ago

A fluid is flowing with an average speed of 1.5 m/s through a tube that has a radius of 2.0 mm and is 18 cm long. The drop in pr

Ymorist [56]

Answer:

μ = 0.038 Pa.s

Explanation:

Given that

V= 1.5 m/s

r= 2 mm

L = 18 cm

If we assume that flow inside the tube is laminar ,then the pressure drop ΔP given as

$\Delta P=\dfrac{32\mu LV^2}{d^2}$

$\mu = \dfrac{\Delta P D^2}{LV^2}$

$\mu = \dfrac{967\times 0.004^2}{0.18\times 1.5^2}\ Pa.s$

μ = 0.038 Pa.s

The viscosity of the fluid = 0.038 Pa.s

5 0

4 years ago

What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of 85 km/h so that the occ

borishaifa [10]

Answer:

k = 5178.8 N/m

Explanation:

As we know that spring mass system will oscillate at angular frequency given as

$\omega = \sqrt{\frac{k}{m}}$

now we have

$\omega = \sqrt{\frac{k}{1200}}$

now the maximum acceleration of the spring block system is at its maximum compression state which is given as

$a = \omega^2 A$

here A= maximum compression of the spring

so here in order to find maximum compression of the spring we will use energy conservation as we know that initial total kinetic energy of the car will convert into spring potential energy

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2$

here we know that

v = 85 km/h

$v = 85 \times\frac{1000}{3600} = 23.61 m/s$

now we have

$(1200)(23.61^2) = kA^2$

$A^2 = \frac{6.68 \times 10^5}{k}$

now from above equation of acceleration we have

$5.0 g = (\frac{k}{m})\sqrt{\frac{6.68 \times 10^5}{k}}$

$5.0(9.81) = \sqrt{k}(0.68)$

$k = 5178.8 N/m$

6 0

3 years ago

A car goes from 80 kn/h to 20 km/h in 0.5 h. What is the acceleration in km/h^2? *

jekas [21]

Answer:

<h2>In physics, acceleration is defined as the rate of change of the velocity of an object with respect to time. That is A = VF - VI /T; </h2><h2></h2><h2>Where A = Acceleration </h2><h2>VF = Final velocity = 20 km </h2><h2>VI = Initial velocity = 80 km </h2><h2>T= Time = 0.5 hr </h2><h2>A = (20 - 80)/ 0.5 = - 60/.5 </h2><h2>A = - 120 Km/ hr^2 </h2><h2></h2><h2>Therefore, the acceleration of the car is -120 km/hr^2. </h2><h2> Because the value is negative, the appropriate name that describe it is deceleration.</h2><h2>I Hope this helps</h2>

3 0

3 years ago