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Lorico [155]
3 years ago
7

You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an el

evator that, conveniently, has a scale. Find the magnitude and direction of the elevator's acceleration if the scale reads
Find the magnitude of the elevator's acceleration if the scale reads 726 N
Find the direction of the elevator's acceleration if the scale reads 726 N
Find the magnitude of the elevator's acceleration if the scale reads 598 N
Find the direction of the elevator's acceleration if the scale reads 598 N
Physics
2 answers:
Lapatulllka [165]3 years ago
7 0

Answer:

a=0.5418\ m.s^{-2} upwards

a=1.283\ m.s^{-2} downwards

Explanation:

Given:

weight of the person, w=688\ N

So, the mass of the person:

m=\frac{w}{g}

m=\frac{688}{9.81}

m=70.132\ kg

  • Now if the apparent weight in the elevator, w_a= 726\ N

<u>Then the difference between the two weights is :</u>

\Delta w=w_a-w

\Delta w=726-688

\Delta w=38\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{38}{70.132}

a=0.5418\ m.s^{-2} upwards, because the normal reaction that due to the weight of the body is increased here.

  • Now if the apparent weight in the elevator, w_a= 598\ N

<u>Then the difference between the two weights is :</u>

\Delta w=w-w_a

\Delta w=688-598

\Delta w=90\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{90}{70.132}

a=1.283\ m.s^{-2} downwards, because the normal reaction that due to the weight of the body is decreased here.

Julli [10]3 years ago
5 0

Answer with Explanation:

We are given that

Weight of person=w=688 N

a.Scale reading=n=726 N

m=\frac{w}{g}=\frac{688}{9.8} m

By using g=9.8m/s^2

Net force=ma=n-w

a=\frac{n-w}{m}=\frac{726-688}{\frac{688}{9.8}}m/s^2

a=0.54 m/s^2

The sign of acceleration(a) is positive.Therefore, the direction of elevator's accelerating is upwards.

b.Scale reading=n=598 N

a=\frac{598-688}{\frac{688}{9.8}}=-1.28m/s^2

a=-1.28 m/s^2

The sign of acceleration(a) is negative.Therefore, the direction of elevator's accelerating is downwards.

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A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti
LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision = 7.668\ ms^-1

Their common speed after the collision = 2.56\ ms^-1

Height achieved by the package of mass m when it rebounds = 0.33\ m

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

where K is Kinetic energy and U is Potential energy.

K= \frac{mv^2}{2} and U= mgh

Considering the fact  K_{initial} = 0\ and U_{final} =0 we will plug out he values of the given terms.

So V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1

Keypoints:

  • Sum of energies and momentum are conserved in all collisions.
  • Sum of KE and PE is also known as Mechanical energy.
  • Only KE is conserved for elastic collision.
  • for elastic collison we have e=1 that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

m_1V_1(i) = (m_1+m_2)V_f where V_1{i} =7.668\ ms^-1.

Plugging the values we have

m\times 7.668 = (3m)\times V_{f}

Cancelling m from both sides and dividing 3 on both sides.

V_f = 2.56\ ms^-1

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic e=1

We have to find the value of V_{f} for m mass.

As here V_{f}=-2.56\ ms^-1 we can use that if both are moving in right ward with 2.56 then there is a  -2.56 velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object 1.

So

V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1

Now using law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh

\frac{v(f1)^2}{2g} = h

h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed =7.668\ ms^-1 for part b we have their common speed =2.56\ ms^-1 and for part c we have the rebound height =0.33\ m.

3 0
3 years ago
Help me to solve it . It’s urgent
Artist 52 [7]

Answer: 0°

Explanation:

Step 1: Squaring the given equation and simplifying it

Let θ be the angle between a and b.

Given: a+b=c

Squaring on both sides:

... (a+b) . (a+b) = c.c

> |a|² + |b|² + 2(a.b) = |c|²

> |a|² + |b|² + 2|a| |b| cos 0 = |c|²

a.b = |a| |b| cos 0]

We are also given;

|a+|b| = |c|

Squaring above equation

> |a|² + |b|² + 2|a| |b| = |c|²

Step 2: Comparing the equations:

Comparing eq( insert: small n)(1) and (2)

We get, cos 0 = 1

> 0 = 0°

Final answer: 0°

[Reminders: every letters in here has an arrow above on it]

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Which of the Jovian planets have rings?
stira [4]
Jupiter , Saturn , Uranus , Neptune .
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A vertical spring withk= 245N/m oscillates with an amplitude of 19.2cm when 0.457kg hangs from it. The mass posses through the e
Makovka662 [10]

Answer:

 y = -19.2 sin (23.15t) cm

Explanation:

The spring mass system is an oscillatory movement that is described by the equation

      y = yo cos (wt + φ)

Let's look for the terms of this equation the amplitude I

     y₀ = 19.2 cm

Angular velocity is

     w = √ (k / m)

     w = √ (245 / 0.457

     w = 23.15 rad / s

The φ phase is determined for the initial condition   t = 0 s ,  the velocity is negative v (0) = -vo

The speed of the equation is obtained by the derivative with respect to time

     v = dy / dt

     v = - y₀ w sin (wt + φ)

For t = 0

     -vo = -yo w sin φ

The angular and linear velocity are related v = w r

      v₀ = w r₀

      v₀ = v₀ sinφ

      sinφ = 1

      φ = sin⁻¹ 1

      φ = π / 4    rad

Let's build the equation

      y = 19.2 cos (23.15 t + π/ 4)

Let's use the trigonometric ratio π/ 4 = 90º

      Cos (a +90) = cos a cos90 - sin a sin sin 90 = 0 - sin a

       y = -19.2 sin (23.15t) cm

8 0
3 years ago
How does changes in distance affect the gravitational pull between two objects? Describe and give one example.
maxonik [38]
The formula is

F_grav = G * m1 * m2 / r^2

G m1 and m2 are going to stay the same once chosen no matter what the distance is. The only thing that will change is the distance.

As the distance increases, the Gravitational Force will decrease. It will decrease by quite a bit.

As the distance decreases, the gravitational force will Increase.

The relationship is inverse. The moon travelling around the earth is one example. The earth travelling around the sun is another.
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