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Mademuasel [1]
2 years ago
14

If a 1.00 kg body has an acceleration of 2.44 m/s2 at 53° to the positive direction of the x axis, then what are (a) the x comp

onent and (b) the y component of the net force on it, and (c) what is the net force in unit-vector notation?
Physics
1 answer:
Ilia_Sergeevich [38]2 years ago
6 0

(a) Fx = 1.464 N

(b) Fy = 1.952 N

(c) F(x, y) = 1.464 i + 1.952 j

Given

Mass = 1kg

Acceleration = 2.44 m/s2

Angle with positive X axis = 53°

As we know

F = ma

By substituting value

F= 1×2.44 N

F= 2.44 N

(a)   Component of force in X direction

Fx = F Cosθ

Fx = 2.44 Cos(53°)

Fx = 2.44 × 0.60 = 1.464 N

(b) Component of force in Y direction

Fy = F Sinθ

Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N

(c) Net force in vector notation

F(x, y) = 1.464 i + 1.952 j

Thus we got net force.

#SPJ4

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DaniilM [7]

Answer: 21 kg x m/s

Momentum can be found using the equation p = mv, or momentum = mass times velocity.

Therefore, to find the momentum of the bowling bowl, you can plug the mass and velocity values into the equation to get p = (7kg)(3m/s)

Solve the equation to get a momentum of 21kg x m/s

I hope I could help :)

7 0
3 years ago
An electrochemical cell has a constant potential of 2 V and provides a steady current of 1 A. How many electrons move from the a
myrzilka [38]

Answer:

n=6.25\times 10^{21}

Explanation:

Potential across the electrochemical cell, V = 2 V

Current, I = 1 A

Time, t = 1000 s

Let n electrons move from the anode to the cathode. We know that electric current is calculated as :

q=I\times t

Also, q = ne

ne=It

n=\dfrac{It}{e}

n=\dfrac{1\times 1000}{1.6\times 10^{-19}}

n=6.25\times 10^{21}

So, 6.25\times 10^{21} electrons move from the anode to the cathode in 1000 seconds.

8 0
4 years ago
A ball rolled 12.0 m [E] in 10.0 s, hit an obstacle, and rolled straight back. After the collision, the ball rolled 8.00 m [W] i
Angelina_Jolie [31]

Answer:

the average velocity of the ball is 5 m/s.

Explanation:

Given;

initial position of the ball, x₁ = 12 m East

final position of the ball, x₂ = 8 m West

initial time of motion, t₁ = 10.0 s

final time of motion, t₂ = 6.0 s

The average velocity is calculated as follows;

v = \frac{\Delta x}{\Delta t} = \frac{x_1 \ - \ x_2}{t_1 \ - \ t_2}

let the Eastward direction be positive

Let the westward direction be negative

\frac{x_1 \ - \ x_2}{t_1 \ - \ t_2} = \frac{12 \ - \ (-8)}{10 -6}= \frac{20}{4} = 5 \ m/s

Therefore, the average velocity of the ball is 5 m/s.

5 0
3 years ago
A car starts from rest and accelerates for 5.2 s with an acceleration of 2.8 m/s 2 . How far does it travel? Answer in units of
Mumz [18]
The final velocity is 5.2^2. The initial velocity is 0 . You are using Kinematic equation 3. (V^2=Vo^2+2a(x)) . (5.2^2=2(2.8)x)
27.04=5.6x
27.04/5.6=5.6/5.6x
4.83=x
The answer is 4.83m:)
5 0
3 years ago
A bass guitar string is 89cm long with a fundamental frequency of 30Hz.
aksik [14]

Answer:

= 53.4 m/s

Explanation:

Velocity of a sound is given by the formula;

Velocity = Frequency × 2 (Length)

Frequency = 30 Hz

Length = 89 cm, equivalent to 0.89 M

Therefore;

Velocity = 30 × 2 ×0.89

              = 53.4 m/s

Thus; the velocity or the speed of sound is 53.4 m/s

5 0
4 years ago
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