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3 years ago

Pls answer my quetion i will give brainlist answer

Physics

1 answer:

allochka39001 [22]3 years ago

7 0

Answer:

here's an explanation but not the answer

please kindly send No or Yes understand.

Explanation:

A 10-N force is applied to push a block across a friction free surface for a displacement of 5.0 m to the right.

See Answer

A 10-N frictional force slows a moving block to a stop after a displacement of 5.0 m to the right.

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A school custodian waxed a gymnasium floor that measured 20 feet by 20 feet. How many square feet did she have to wax?

Vadim26 [7]

We have to calculate how many square feet a school custodian has to wax and to choose the correct answer. A gymnasium floor is in the shape of the square ( 20 by 20 feet ). The area of the square is : A = L^2. For L = 20 ft, A = 20^2 = 20 * 20 = 400 ft^2. Answer: D. 400 square feet.

5 0

4 years ago

A mysterious rocket-propelled object of mass 47.5kg is initially at rest in the middle of the horizontal, frictionless surface o

NARA [144]

Answer:

Distance d=8.63m

Explanation:

mass of object m=47.5 kg

Initial velocity vi=0 m/s

Magnitude of force on object F(t)=(17.0 N/s)t

To find

Distance

Solution

The rocket accelerates due to variable force so we apply Newtons second law but the acceleration will not be constant because the force is not constant

We use ax=Fx/m to find acceleration but it must be integrate to find velocity and then the distance that rocket travels

$ax=Fx/m\\gives\\ax(t)=(17.0N/s)t/(47.5kg)\\ax(t)=(0.35789m/s^{3})t\\ As\\Vx(t)=V_{o}+\int\limits^{}_{} { \lim_{0 \to \ t} ax(t) } \, dt=(0.178945m/s^{3} )t^{2}\\ X-X_{o}=\int\limits^{}_{} { \lim_{0 \to \ t} v(t) } \, dt=(0.05964m/s^{3})t^{3}\\ as\\t=5.25s\\So\\X-X_{o}=(0.05964m/s^{3})(5.25s)^{3}\\X-X_{o}=8.63m$

Distance d=8.63m

8 0

3 years ago

A skateboarder rolls off a horizontal ledge that is 1.32m high and lands 1.88m from the base of the ledge. What was the initial

NemiM [27]

Answer:

Velocity = 3.622 m/s

Explanation:

Given:

Starting height = 1.32 m

Distance = 1.88 m

Find:

Velocity

Computation:

Using 2nd equation of motion.

s = ut + [1/2]gt² , where g = 9.8 m/s² and u = 0

1.32 = [1/2][9.8]t²

Time (t) = 0.519 sec

We know that

Velocity = Distance / Time

Velocity = 1.88 / 0.519

Velocity = 3.622 m/s

4 0

4 years ago

The circuit in Fig. P4.23 utilizes three identical diodes having IS = 10−14 A. Find the value of the current I required to obtai

Sliva [168]

Answer:

$v = \frac{2 V}{3}= 0.667 v$

Since we have identical diodes we can use the equation:

$I_D =I= I_S e^{\frac{V_D}{V_T}}$

And replacing we have: $I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA$

Since we know that 1 mA is drawn away from the output then the real value for I would be

$I_D = I = 3.86 mA -1 mA= 2.86 mA$

And for this case the value for $v_D$ would be:

$V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V$

And the output votage on this case would be:

$V = 3 V_D = 3 *0.660 V = 1.98 V$

And the net change in the output voltage would be:

$\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V$

Explanation:

For this case we have the figure attached illustrating the problem

We know that the equation for the current in a diode id given by:

$I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}$

For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode $v_1 + v_2 + v_3= 2$ and each voltage is the same v for each diode, so then:

$v = \frac{2 V}{3}= 0.667 v$

Since we have identical diodes we can use the equation:

$I_D =I= I_S e^{\frac{V_D}{V_T}}$

And replacing we have:

$I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA$

Since we know that 1 mA is drawn away from the output then the real value for I would be

$I_D = I = 3.86 mA -1 mA= 2.86 mA$

And for this case the value for $v_D$ would be:

$V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V$

And the output votage on this case would be:

$V = 3 V_D = 3 *0.660 V = 1.98 V$

And the net change in the output voltage would be:

$\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V$

5 0

3 years ago

What’s the velocity of a bike ride taking 45 mins over a distance of 4km

Y_Kistochka [10]

The speed of the ride is

Speed = (distance covered) / (time to cover the distance)

Speed = (4,000 m) / (45 min)

Speed = 88.89 m/s

Speed = 1.481 m/s

Speed = 5.333 km/hr

There's not enough information given in the question to calculate the velocity of the ride. For example ...

-- If the ride was completely in a straight line, then the velocity would be exactly equal to the speed.

-- If the rider went 2km from his house and then 2km back home again, his velocity for the whole ride would be zero.

We just don't know from the information given.

8 0

3 years ago

Pls answer my quetion i will give brainlist answer​

Pls answer my quetion i will give brainlist answer