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lesantik [10]
3 years ago
7

Vector A⃗ points in the negative y direction and has a magnitude of 5 km. Vector B⃗ has a magnitude of 15 km and points in the p

ositive x direction. Use components to find the magnitude of
a) A+B
b) A-B
c) B-A
Physics
1 answer:
Harlamova29_29 [7]3 years ago
7 0
C
Because it’s B-A if u reared the question u will understand
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A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

3 0
3 years ago
URGENT!!! NEED ANSWER ASAP Students are completing a table about a particular subatomic particle that helps make up an atom. The
Contact [7]
Hmm, I will come back to this one just to help. :)
3 0
2 years ago
Read 3 more answers
Calculate the force exerted on the wall assuming that force is horizontal and using the data in the schematic representation of
Jlenok [28]

Answer:

1.93 x 10∧3 N

Explanation:

The picture attached shows the calculation

8 0
2 years ago
An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same
Hatshy [7]

Answer:

v_2=\sqrt{2}v_1

Explanation:

The velocity v₁ can be calculated with the kinematic formula:

v_1^{2} =v_0^{2} +2gh

Since the object is initially at rest, v₁ becomes:

v_1=\sqrt{2gh}

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

v_2^{2}=v_1^{2} +2gh

Substituting v₁ in this expression and solving for v₂, we get:

v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}

Now, dividing v₂ over v₁, we get the expression:

\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\   \\\implies v_2=\sqrt{2}v_1

It means that v₂ is √2 times v₁.

4 0
3 years ago
What causes breakers?
nata0808 [166]

Answer:

nreaker

Explanation:

A switch that automatically interrupts or shuts off an electric current at the first indication of a overload

3 0
3 years ago
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