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myrzilka [38]
3 years ago
6

At t=0, a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xy plane with a con

stant acceleration of (2.0i - 4.0j) m/s2. At the instant the x coordinate of the particle is 15 m, what is the speed of the particle?
a. 10 m/s
b. 16 m/s
c. 12 m/s
d. 14 m/s
e. 26 m/s
Physics
1 answer:
fiasKO [112]3 years ago
5 0

Answer:

e.26m/s

Explanation:

Vf=Vi+at      (1)

Vf=9j+(2i-4j)t

X= X₀+at

now, in the i direction

15=O+2t or t=7.5 when x position is 15

Lets put that into the (1) equation, solve for Vf.

 Vf=9j+(2i-4j)7.5

Vf= 15i - 21j

Speed= \sqrt{xcomponent^2 + ycomponent^2}

Vf= 25.8 m/s

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A sprinter accelerates from rest to 10.0 m/s in 1.28 s . Part A Part complete What is her acceleration in m/s2? a a = 7.81 m/s2
Mashutka [201]

Explanation:

It is given that,

Initial speed of sprinter, u = 0

Final speed of sprinter, v = 10 m/s

Time taken, t = 1.28 s

a. We need to find the acceleration of sprinter. It can be calculated using first equation of motion as :

a=\dfrac{v-u}{t}

a=\dfrac{10\ m/s}{1.28\ s}

a=7.81\ m/s^2

b. Final speed of the sprinter, v = 36 km/h

Time, t = 0.000355 h

Acceleration, a=\dfrac{36}{0.000355}

a=101408.45\ km/h^2

Hence, this is the required solution.

3 0
4 years ago
A number is written in scientific notation when it is the product of a number less than 10 but greater than or equal to 1 and a
Bess [88]

Answer:

Mass = 1.99 * 10^{30} kg

Explanation:

Given

Mass = 1,990,000,000,000,000,000,000,000,000,000\ kg

Required

Rewrite using scientific notation

The format of a number in scientific notation is

Digit = a * 10^n

Where 1 \geq\ a\geq\ 10

So the given parameter can be rewritten as

Mass = 1.99 * 1,000,000,000,000,000,000,000,000,000,000\ kg

Express as a power of 10

Mass = 1.99 * 10^{30} kg

Hence, the equivalent of the mass of the sun in scientific notation is:

Mass = 1.99 * 10^{30} kg

7 0
3 years ago
1) the strength of an electromagnet can be increased by
miss Akunina [59]
Using coils of fewer turns on the electromagnet
5 0
3 years ago
A technician wishes to determine the wavelength of the light in a laser beam. To do so, she directs the beam toward a partition
Nana76 [90]

Answer:

λ = 5.656 x 10⁻⁷ m = 565.6 nm

Explanation:

Using the formula of fringe spacing from the Young's Double Slit experiment, which is given as follows:

\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x\ d}{L}

where,

λ = wavelength = ?

Δx = fringe spacing = 1.6 cm = 0.016 m

L = Distance between slits and screen = 4.95 m

d = slit separation = 0.175 mm = 0.000175 m

Therefore,

\lambda = \frac{(0.016\ m)(0.000175\ m)}{4.95\ m}\\\\

<u>λ = 5.656 x 10⁻⁷ m = 565.6 nm</u>

6 0
3 years ago
Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an ar
den301095 [7]

Answer:

electric field   Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

 a) Let's write the electric field for each charge and the total field

       E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

       Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

     (1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

       

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

   Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

3 0
3 years ago
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