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GrogVix [38]
3 years ago
13

You hold a 0.125 kg glider A and a 0.500 kg glider B at rest on an air track with a compressed spring of negligible mass between

them. When you release the gliders, the spring pushes them apart so that they move in opposite directions. When glider A has moved 0.960 m to the left from its starting position, how far to the right from its starting position has glider B moved?
Physics
1 answer:
o-na [289]3 years ago
8 0

Answer:

Magnitude |VB| = 0.32 m/s

and

Direction is to the left

Explanation:

given data

glider A mass = 0.125 kg

glider B mass = 0.500 kg

glider A move = 0.960 m to the left

solution

we use here Conservation of momentum  that is

M1 × U1 + M2 × U2 =M1 × V1 + M2 × V2    ........................1

put here value and we get

0.125 × 0 + 0.500 × 0 = 0.125 × 0.960 + 0.375 × VB

VB = −0.32  m/s

Magnitude |VB| = 0.32 m/s

and

Direction is to the left

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A cube of ice at an initial temperature of -15.00°C weighing 12.5 g total is placed in 85.0 g of water at an initial temperature
Leona [35]

<u>Answer:</u> The specific heat of ice is 2.11 J/g°C

<u>Explanation:</u>

When ice is mixed with water, the amount of heat released by water will be equal to the amount of heat absorbed by ice.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of ice = 12.5 g

m_2 = mass of water = 85.0 g

T_{final} = final temperature = 22.24°C

T_1 = initial temperature of ice = -15.00°C

T_2 = initial temperature of water = 25.00°C

c_1 = specific heat of ice = ?

c_2 = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

12.5\times c_1\times (22.24-(-15))=-[85.0\times 4.186\times (22.24-25)]

c_1=2.11J/g^oC

Hence, the specific heat of ice is 2.11 J/g°C

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4 years ago
A ball is rolling across the floor at a constant speed. What will happen to the ball if it is exposed to an unbalanced force in
KATRIN_1 [288]

Answer:

B. it's speed will increase

7 0
3 years ago
Read 2 more answers
"A 75 kg swimmer dives horizontally off a 500 kg raft. If the diver's speed immediately after leaving the raft is 4 m/s, what is
svet-max [94.6K]

Answer:

0.6 m/s

Explanation:

Let the raft speed be v.

Given:

Mass of the swimmer (m) = 75 kg

Mass of the raft (M) = 500 kg

Speed of swimmer after leaving the raft (u) = 4 m/s

Now, the given situation can be considered a problem of conservation of total momentum before and after collision.

Momentum is the product of mass and velocity.

Here, the swimmer and raft are the bodies in collision.

So, before the collision, both the bodies were at rest. So, initial momentum is 0. Now, from conservation of momentum, the final momentum of the system must be 0 after the collision. Therefore,

Final Momentum = 0

mu+Mv=0\\\\Mv=-mu\\\\v=-\dfrac{mu}{M}

Plug in the given values and solve for 'v'. This gives,

v=-\frac{75\times 4}{500}\\\\v=-\frac{300}{500}=-0.6\ m/s

The final velocity of the raft is -0.6 m/s. Now, speed is the magnitude of velocity.

Therefore, the corresponding raft speed is 0.6 m/s.

7 0
4 years ago
a wall of glass 2cm thickhas inside temperature of 30°C,outside temperature of15°C.how much heat is flowing through the glass(k=
kenny6666 [7]

Answer:46.5

Explanation:is the topical formula of F= 30x+15

5 0
4 years ago
An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric
ELEN [110]

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ =  -ε²/(R + r)²

cancelling out the common variables

2R =  R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

8 0
4 years ago
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