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Tema [17]
2 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]2 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

You might be interested in
A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of
lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

8 0
2 years ago
Read 2 more answers
A 330.-ohm resistor is connected to a 5.00-volt battery. What is the current through the resistor?
Gelneren [198K]

I = 0.0152 A = 15.2 mA

Explanation:

Using Ohm's law,

V = IR or

I = V/R

= (5.00 V)/(330. ohms)

= 0.0152 A

= 15.2 mA

8 0
2 years ago
Read 2 more answers
In an LC circuit containing a 40-mH ideal inductor and a 1.2-mF capacitor, the maximum charge on the capacitor is 45mC during os
GalinKa [24]

Answer:

E) 6.5 A

Explanation:

Given that

L = 40 m H

C= 1.2 m F

Maximum charge on capacitor ,Q= 45 m C

The maximum current I given as

I = Q.ω

ω =angular frequency

\omega=\dfrac{1}{\sqrt{CL}}

By putting the values

\omega=\dfrac{1}{\sqrt{CL}}

\omega=\dfrac{1}{\sqrt{40\times 10^{-3}\times 1.2 \times 10^{-3}}}

ω  = 144.33 rad⁻¹

Maximum current

I = 45 x 10⁻³ x 144.33  A

I= 6.49 A

I = 6.5 A

E) 6.5 A

3 0
3 years ago
Calculate Vector component in Y if the hypotenuse is 32 and angle is 45
Lerok [7]

Answer:

The correct option is;

c. 22.6

Explanation:

The given parameters are;

The hypotenuse of the vector = 32

The angle of the vector = 45°

Therefore, the vector component in the y-axis is given as follows;

v_y = v \times sin(\theta)

Substituting the values from the question gives;

v_y = 32 \times sin(45^{\circ}) \approx 22.6

The vector component in the y-axis, v_y, is approximately 22.6.

8 0
2 years ago
!!HELP HELP!!
borishaifa [10]

Answer:

7.5s

Explanation:

Given parameters:

Velocity  = 30m/s

Deceleration  = 4m/s²

Unknown:

Time it takes for the car to come to complete rest  = ?

Solution:

 To solve this problem, we use the kinematics expression below:

        v  = u + at

 Since this is a deceleration

         v  = u - at

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

         v - u  = -at

         0  - 30  = -4 x t

              -30  = -4t

               t  = 7.5s

6 0
2 years ago
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