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Irina18 [472]
3 years ago
9

Blythe and Geoff are ice-skating together. Blythe has a mass of 40 kg and Geoff has a mass of 79 kg. Blythe pushes Geoff in the

chest when both are at rest, causing him to move away at a speed of 5 m/s.

Physics
2 answers:
alexandr1967 [171]3 years ago
6 0

Answer:

Speed of Blythe is 9.8 m/s.

Explanation:

Mass of Blythe =40 kg

Mass of Geoff = 79 kg

Speed = 5 m/s

Suppose, we determine the Blythe's speed after she pushes Geoff

Since, initial momentum is zero final momentum should be zero.

Using momentum of conservation

m_{B}v_{B}+m_{G}v_{G}=0

v_{B}=-\dfrac{m_{G}v_{G}}{m_{B}}

Put the value into the formula

v_{B}=-\dfrac{79\times5}{40}

v_{B}=-9.8\ m/s

Negative sign shows that he move in direction opposite to Geoff.

Hence, Speed of Blythe is 9.8 m/s.

Deffense [45]3 years ago
4 0
9.9 m/s

calculations below 

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considere que o calor específico de um material presente nas cinzas seja c=0,8j/gc. Supondo que esse material entre na turbina a
drek231 [11]

Answer:

3120J

Explanation:

Given parameters:

C  = Specific heat capacity  = 0.8J/g°C

Initial temperature  = 20°C

Mass given   = 5g

Final temperature  = 800°C

Unknown:

Energy given to the mass  = ?

Solution:

To find the energy given to the mass, let us simply use the expression below:

          H   =   m   c   ΔT

H is the unknown, the energy supplied

m is the mass of the substance

c is the specific heat capacity

ΔT is the change in temperature

Input the variables;

            H    = 5  x   0.8    x    (800 - 20)  = 3120J

7 0
3 years ago
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal
kirza4 [7]

At t =0, the velocity of A is greater than the velocity of B.

We are told in the question that the spacecrafts fly parallel to each other and that for the both  spacecrafts, the velocities are described as follows;

A: vA (t) = ť^2 – 5t + 20

B: vB (t) = t^2+ 3t + 10

Given that t = 0 in both cases;

vA (0) = 0^2 – 5(0) + 20

vA = 20 m/s

For vB

vB (0) = 0^2+ 3(0) + 10

vB = 10 m/s

We can see that at t =0, the velocity of A is greater than the velocity of B.

Learn more: brainly.com/question/24857760

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?

3 0
3 years ago
A proton and an electron are released from rest, with only the electrostatic force acting. Which of the following statements mus
lbvjy [14]

Answer:

their electrical potential energy decreases. True, because of the negative sign as the distance decreases, it becomes more negative.

The kinetic energy increases. True, the force is attractive as the distance decreases the force increases, so acceleration and therefore the speed

Explanation:

The only force between the proton and the electron is electric

      Fe = k q1q2 / r2

      Fe = - k e2 / r2

We can see that it is an attractive force (negative sign)

The electric power energy is

       U = k q1 q2 / r

       U = -k e2 / r

 

The kinetic energy is

       K = ½ m v2

With the expressions for each term we can analyze the sentences :

Their electric potential energy increases. False,

their electrical potential energy decreases. True, because of the negative sign as the distance decreases, it becomes more negative.

The kinetic energy increases. True, the force is attractive as the distance decreases the force increases, so acceleration and therefore the speed

Kinetic energy decreases False

The acceleration decreases.  False, as the force increases so does the acceleration

7 0
3 years ago
block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8
dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: \sqrt{800} \,\,\,N

Explanation:

First try to write all forces in  vector component form:

The force F1 acting at 45 degrees would have multiplication factors of \frac{\sqrt{2} }{2} on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is    F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

y-component of F1 is    F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

y-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is:     v_x=4\,*\,2=8\,m/s

y-component of the velocity is:     v_y=4\,*\,2=8\,m/s

7 0
3 years ago
Some liquid is poured into a burrete so that it reads 14cm³.50 drops were run each of volume 0.1cm³ .
nika2105 [10]

Given :

Liquid is poured into a burrete so that it reads 14cm³.

50 drops were run each of volume 0.1cm³ .

To Find :

The volume of liquid in burrete after 50 drops.

Solution :

Volume of each drop, v = 0.1 cm³.

Initial volume in burrete, V = 14 cm³.

Now, volume left after droping 50 drops are :

L = V - 50v\\\\L = 14 - 50\times 0.1 \ cm^3 \\\\L = 9 \ cm^3

Therefore, the volume left in burrete is 9 cm³ .

5 0
3 years ago
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