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Tema [17]
3 years ago
9

A child uses a rubber band to launch a bottle cap at an angle of 37.0° above the horizontal. The cap travels a horizontal distan

ce of 1.30 m in 1.20 s. What was the initial speed of the bottle cap, just after leaving the rubber band, in m/s
Physics
1 answer:
zavuch27 [327]3 years ago
3 0

Answer:

Initial velocity will be 1.356 m/sec      

Explanation:

Let the initial speed = u

Angle at which rubber band is launched = 37°

Horizontal component of initial velocity u_x=ucos\Theta =ucos37^{\circ}=0.7986u

Time is given as t = 1.20 sec

Distance in horizontal direction = 1.30 m

We know that distance = speed × time

So time t=\frac{distance}{speed}

1.20=\frac{1.3}{0.7986u}

u=1.356m/sec

So initial velocity will be 1.356 m/sec

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arlik [135]

Motion with Constant Acceleration

Constant acceleration occurs when an object's velocity changes by an equal amount in every equal time period.

There are five frequently used formulas for motion in a straight line with constant acceleration. The formulas are given in terms of the initial velocity u, the final velocity v, the displacement (position) x, the acceleration a and the time elapsed t.

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3 years ago
two forces act concurrently on an object on a horizontal frictionless surface. their resultatn force has the largest magnitude w
Gennadij [26K]

Answer:

54

Explanation:4

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5 0
2 years ago
A torque of 50 N*m acts on a wheel of moment of inertia 25 kg*m^2 for 4 s and then is removed.
motikmotik
You have to find the calculate<span> the circumference first then you can just multiply the diameter by π, which is about 3.142. That gives you the distance for each </span>revolution<span>. Then you can multiply by the </span>number of revolutions<span> per minute.

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3 0
3 years ago
A well lagged copper calorimeter of mas 120g contains 70g of water and 10g ice both at 0°C . Dry steam at 100°C is passed in unt
Lina20 [59]

Answer:

7.6 g

Explanation:

"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.

The heat gained by the copper, water, and ice = the heat lost by the steam

Heat gained by the copper:

q = mCΔT

q = (120 g) (0.40 J/g/K) (40°C − 0°C)

q = 1920 J

Heat gained by the water:

q = mCΔT

q = (70 g) (4.2 J/g/K) (40°C − 0°C)

q = 11760 J

Heat gained by the ice:

q = mL + mCΔT

q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)

q = 4880 J

Heat lost by the steam:

q = mL + mCΔT

q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)

q = 2452 J/g m

Plugging the values into the equation:

1920 J + 11760 J + 4880 J = 2452 J/g m

18560 J = 2452 J/g m

m = 7.6 g

7 0
3 years ago
Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly
loris [4]

Answer: 36.86\°

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: d=2.4 m

Ferdinand's initial velocity: V_{o}=5 m/s

Time it takes a jump: t=0.6 s

We need to find the angle \theta at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

V_{x}=\frac{d}{t} (1)

V_{x}=\frac{2.4 m}{0.6 s} (2)

V_{x}=4 m/s (3)

On the other hand, the x-component of the velocity is expressed as:

V_{x}=V_{o}cos\theta (4)

Substituting (3) in (4):

4 m/s=5 m/s cos\theta (5)

Clearing \theta:

\theta=cos^{-1} (\frac{4 m/s}{5 m/s})

\theta=36.86\° This is the angle at which Ferdinand the frog jumps between lily pads

4 0
3 years ago
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