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Anna [14]
3 years ago
15

As the launch force increase the launch velocity will

Physics
1 answer:
FrozenT [24]3 years ago
5 0

Answer:

As the launch force increase the launch velocity will

<em><u>Increase</u></em>

The reason for your answer to number six is because

<em><u>There is a direct relationship between force and acceleration.</u></em>

<em><u /></em>

Explanation:

<em>It is known all over the place that, there is a direct relationship between Force and acceleration of an object leading to an increase in force being directly proportional to the increase in the acceleration of the given object and vice versa.</em>

You might be interested in
. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in
kupik [55]

Answer:

(c) 16 m/s²

Explanation:

The position is r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}.

The velocity is the first time-derivative of <em>r(t).</em>

<em />v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}<em />

The acceleration is the first time-derivative of the velocity.

a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}

Since <em>a(t)</em> does not have the variable <em>t</em>, it is constant. Hence, at any time,

a = -16\hat{j}

Its magnitude is 16 m/s².

4 0
3 years ago
How would you write the number 6,500,000,000 in scientific notation?
photoshop1234 [79]
Scientific form = 6.5 x 109.
8 0
3 years ago
A stone that starts at rest free falls for 7.0 s. How far does the stone fall in this time?
vredina [299]

Gravitational acceleration is approx 9.8 m/s
Time is 7s

a=9.8 m/s
t=7s

a = d/t^2

therefore:

d = a * t^2

d = 9.8 * 7^2

d = 9.8 * 49

d = 480.2 [m]

7 0
3 years ago
A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr
valina [46]
The  spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
T=2 t = 2 \cdot 0.100 s = 0.200 s
Which means that the frequency is
f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz
and the angular frequency is
\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s

In a spring-mass system, the maximum velocity of the object is given by
v_{max} = A \omega
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s
6 0
3 years ago
A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif
Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c)  w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)  

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

                    μ = N*I*A

Where,

           N: The number of turns

           I : Current in coil

           A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

                    μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

                    μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

             vec ( τi ) = vec ( μi ) x vec ( B )

              = 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

              = ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

          \left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right]  = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\

- The initial torque ( τi ) is written as follows:

           vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

           

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

          Ui = - vec ( μi ) . vec ( B )

          Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

          Ui = - [(-0.000605556 + 0.00011)]

          Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

          Uf = - vec ( μf ) . vec ( B )

          Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

          Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

          ΔU = Uf - Ui

          ΔU = -0.000252315 - 0.000495556  

          ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

                Ui + Eki = Uf + Ekf

Where,

             Eki : The initial kinetic energy ( initially at rest ) = 0

             Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.    

               -ΔU = Ekf

                0.5*T*w^2 = -ΔU

                w^2 = -ΔU*2 / T

Where,

                w: The angular speed at second position

               w = √(0.000747871*2 / 6.50×10^−7)

              w = 47.97 rad / s

6 0
3 years ago
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