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Tju [1.3M]
2 years ago
4

Wha is the length of the x-component of the vector plotted below?

Physics
2 answers:
mylen [45]2 years ago
6 0

B. 1

Because the x coordinate of the end point is 1 .

Point is (1,5)

so Answer is B .

Yanka [14]2 years ago
4 0

Answer:

Correct Answer is option B that is 1

Explanation:

when a vector is break into its parts, those parts are called its components.  in this question length of  vector along x-axis is 1, and length of vector along y-axis is 5. Therefore X component (also known as horizontal component) of vector is 1 and Y Component(Alson Known as Vertical component) is 5.

Hope this answer will help you

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A carmaker has designed a car that can reach a maximum acceleration of 12 meters/second2. The car’s mass is 1,515 kilograms. Ass
Vlada [557]
1) 15 / 12 = 1.25 ratio
2) to increase acceleration  1.25 times (with same F, or same engine) you have to lower mass 1.25 times
3) 1515/1.25 = 1212 kg

choose A

6 0
2 years ago
At the circus, a 100-kilogram clown is fired 15 meters per second from a 500-kilogram cannon. What is the recoil speed of the ca
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Recoil speed= 3m/s, method shown in photo

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2 years ago
A uniform cylindrical steel wire (density: 7.8 x 103 kg/m3), 58.0 cm long and 1.34 mm in diameter, is fixed at both ends. To wha
lbvjy [14]

Answer:

T= 354.65 N

Explanation:

Given that

ρ= 7800 kg/m³

L= 58 cm

d=1.34 mm

f= 311 Hz

T= Tension

Speed of wave ,V

V = f λ      

V = f L

V= 311 x 0.58 m/s

V=180.38 m/s

Area of cross sectional

A= πr² mm²

A= 3.14 x 0.67² mm²

A=1.41 mm²

Mass = Density x Volume

m=7800\times 1.41\times 10^{-6}\ kg/m

m=0.0109 kg/m

Tension ,T

T=m V^2

T= 0.0109 x 180.38² N

T= 354.65 N

     

3 0
2 years ago
A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a
o-na [289]

Answer:395.6 m/s

Explanation:

Given

mass of bullet m=5 gm

mass of wood block M=1 kg

Length of string L=2 m

Center of mass rises to an height of 0.38 cm

initial velocity of bullet u=450 m/s

let v_1 and v_2 be the velocity of bullet and block after collision

Conserving momentum

mu=mv_1+Mv_2 -------------1

Now after the collision block rises to an height of 0.38 cm

Conserving Energy for block

kinetic energy of block at bottom=Gain in Potential Energy

\frac{Mv_2^2}{2}=Mgh_{cm}

v_2=\sqrt{2gh_{cm}}

v_2=\sqrt{2\times 9.8\times 0.38}

v_2=0.272 m/s

substitute the value of v_2 in equation 1

5\times 450=5\times v_1+1000\times 0.272

v_1=395.6 m/s

4 0
2 years ago
Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a
serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

\omega = 1.25 rad/s \rightarrow The angular speed

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r = 4.65 m \rightarrow The distance

The relation between the linear velocity and angular velocity is

v = r\omega

Where,

r = Radius

\omega = Angular velocity

At the same time we have that the centripetal acceleration is

a_c = \frac{v^2}{r}

a_c = \frac{(r\omega)^2}{r}

a_c = \frac{r^2\omega^2}{r}

a_c = r \omega^2

a_c = (4.65 )(1.25 rad/s)^2

a_c = 7.265625 m/s^2

Now the tangential acceleration is given as,

a_t = \alpha r

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\alpha = Angular acceleration

r = Radius

\alpha = (0.745)(4.65)

\alpha = 3.46425 m/s^2

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

|a| = \sqrt{a_c^2+a_t^2}

|a| = \sqrt{(7.265625)^2+(3.46425)^2}

|a| = 8.049 m/s^2 \approx 8.05 m/s2

Therefore the correct answer is C.

7 0
2 years ago
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