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Vladimir79 [104]
2 years ago
4

A dog runs in a straight line along the x-axis to retrieve a thrown toy. The dog starts from rest at x = 0 at time t = 0, runs t

o x = +15.0 m and comes momentarily to rest there at t = 4.0 s, turns around and runs back past x = 0 at time t = 7.5 s, and finally comes to rest again at x = -10.0 m at t = 10.0 s. In order to start and stop moving at appropriate points, the dog accelerates smoothly as necessary.In answering these questions, it may help if you draw a sketch of the situation, marking the dog's position at various times.What is the dog's displacement (in m) for the entire trip?
Physics
1 answer:
Scorpion4ik [409]2 years ago
5 0

Answer:

the displacement is -10m or 10m on the (-)X axis.

Explanation:

We know that the displacement is a vectorial value,it has a magnitude and a direction.

So in order to obtain the total displacement we have to sum the distance on each trame taking the direction into consideration, we will refer to the right side of the X axis as positive.

On the first stretch, the dog ends up at 15m on the right side, then came back to  x=0m and kept going 10 m more.

d= 15m -15m -10m=-10m

so the total displacement is -10m or 10m on the -X direction.

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A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle
denpristay [2]

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/s^{2} = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = \frac{8.04N}{70.2Kg} = 0.1130801680m/s^2.

and velocity of V = 0.1130801680m/s.

at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

4 0
2 years ago
1) My 14V car battery could be used to charge my laptop, but I need to use an inverter to first convert it to a standard 120V. T
LenaWriter [7]

Answer:

1) Charge chord resistance is 75 Ω

2) Charge chord resistance is 6.33 Ω

Explanation:

1) To answer the question, we note that the the formula voltage is found as follows;

V = IR

Therefore,

R = \frac{V}{I} =  \frac{120}{1.6} = 75  \, \Omega

2) Where the voltage, V = 19.5 V and the current, I = 3.33 A, we have;

Initial resistance R₁ = 19.5 V/(3.33 A) = 5.86 Ω

However, to reduce the current to 1.6 A, we have;

R_T = \frac{19.5}{1.6} = 12.1875 \ \Omega

Therefore, where the resistance is found by the sum of the total resistance we have;

R_T = R₁ + Charge chord resistance

∴ 12.1875 = 5.86 + Charge chord resistance

Hence, charge chord resistance = 6.33 Ω

8 0
2 years ago
A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp
liq [111]

Answer:

W_2=\sqrt{\frac{3}{5} }W_1

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

I_1 =\frac{2}{5}MR^2

K_1 = \frac{1}{2}IW_1^2

So, replacing, we get that:

K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

At the same way, the moment of inertia and kinetic energy for second ball is:

I_2 =\frac{2}{3}MR^2

K_2 = \frac{1}{2}IW_2^2

So:

K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2

Then, K_2 is equal to K_1, so:

K_2 = K_1

\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2

\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2

Finally, solving for W_2, we get:

W_2=\sqrt{\frac{3}{5} }W_1

5 0
2 years ago
Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.
lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = (6\times0.1)kg/min=0.6kg/min

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = \frac{dx}{dt}

So, the rate of change of salt in the tank can be given by the following equation,

1000\frac{dx}{dt} =0.6-6x

or, \int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt

or, T = 693.147 min      (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0
2 years ago
A batter swings at a baseball. The action force is the bat hitting the ball with a force of 5N. What is the reaction force?
omeli [17]
The ball hitting the bat
6 0
2 years ago
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