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Aleksandr [31]
3 years ago
6

Trinity and a penguin get into an argument. The penguin gets angry and attacks her. Its mass is 5 kg and slides at 8 m/s, and Tr

inity is 40 kg and at rest. They collide and bounce apart. After the collision, the penguin is moving at 0 m/s. How fast is Trinity moving after the collision?
Physics
1 answer:
Mice21 [21]3 years ago
4 0

Answer:

velocity of Trinity = 1m/s

Explanation:

Conservation of Momentum equation

m1u1+m2u2=m1v1+m2v2.

m1 = mass of penguin

u1 = initial velocity of penguin

v1 = final velocity of the penguin

m2 = mass of Trinity

u2 = initial velocity of Trinity

v2 = final velocity of trinity

5kg • 8m/s + 40kg • 0m/s = 5kg • 0m/s + 40kg • v2

40kg•m/s = 40kg • v2

divide each side by 40kg

1m/s = v2

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Conduction. is a transfer of energy by direct contact
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A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.
Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0
3 years ago
Roseanne and Jackie were asked to design and create a paper airplane that would stay in the air for as long as possible. They be
jekas [21]

Answer:

the planes times in the air

Explanation:

When a paper airplane is thown the path it takes is not the same every time. So, measuring the distance between the points where the airplane took off and the point where is landed would not show the time the plane was in the air.

The weight also does not determine the time the plane was in the air.

So, the planes air time recorded with the stopwatch gives the time the plane was in the air and will indicate the best plane.

6 0
3 years ago
"An alpha particle (He2+) with a velocity of 2.6x106m/scrosses a uniform magnetic field at an angle of 37.0°to the field lines a
sergey [27]

Answer:

B = 5.59x10⁹ T

Explanation:

The magnetic force (F), on a the alpha particle with charge (q) that is moving at velocity (v) as the cross product of the velocity and magnetic field (B) is:

F = qvBsin(\theta)

<u>We have:</u>

F = 1.4x10⁻³ N

v = 2.6x10⁶ m/s

θ = 37.0°

q = 2*p = 2*1.6x10⁻¹⁹ C

Hence, the strength of the magnetic field is:

B = \frac{F}{qvsin(\theta)} = \frac{1.4 \cdot 10^{-3}}{1.6 \cdot 10^{-19} C*2.6 \cdot 10^{6}*sin(37)} = 5.59 \cdot 10^{9} T

Therefore, the strength of the magnetic field is 5.59x10⁹ T.

I hope it helps you!

7 0
3 years ago
A 6 ft tall person walks away from a 10 ft lamppost at a constant rate of 5 ft/s. What is the rate (in ft/s) that the tip of the
nalin [4]

Answer:

12.5 ft/s

Explanation:

Height of person = 6 ft

height of lamp post = 10 ft

According to the question,

dx / dt = 5 ft/s

Let the rate of tip of the shadow moves away is dy/dt.

According to the diagram

10 / y = 6 / (y - x)

10 y - 10 x = 6 y

y = 2.5 x

Differentiate both sides with respect to t.

dy / dt = 2.5 dx / dt

dy / dt = 2.5 (5) = 12.5 ft /s

8 0
4 years ago
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