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2 years ago

An object is moving east, and its velocity changes from 65 m/s to 25 m/s in 10 seconds Which describes the acceleration?

Physics

1 answer:

Rasek [7]2 years ago

8 0

Answer:

4 m/s in negative acceleration

Explanation:

Acceleration = V- U/t

Where V is the final velocity

U is the initial velocity and t is the time given.

U = 65 m/s

V= 25 m/s

T= 10 seconds

Acceleration= (25m/s - 65m/s)÷10secs

= - 40/10

= -4m/s^2

Hence, it has a negative acceleration.

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Velocity: You leave on a 549-mi trip in order to attend a meeting that will start 10.8 h after you begin your trip. Along the wa

rewona [7]

Answer:

1.55 h

Explanation:

Let the time to spend over dinner be t.

Since I need to spend 10.8 h for the whole trip, then I have 10.8 - t hours left to drive.

Speed = distance/time

$v = \dfrac{d}{10.8-t}$

$d = v(10.8-t)$

The distance is 549 and maximum speed is 65 mi/h.

$549 = 65(10.8 - t)$

$10.8-t = 8.45$

$t = 10.8 - 8.45 = 1.55 \text{ h}$

5 0

2 years ago

A water tank is filled with water. What will be the pressure of the water when the level of the water is 6m?

Luden [163]

Answer:

pressure = density x g x height

= 1000 x 10 x 6 Pascal

=60000 Pascal

OR 60 kP

3 0

2 years ago

Read 2 more answers

A deuteron, 21H, is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a n

Naddik [55]

Answer:

1917723.40119 m/s

Explanation:

m = Mass of deuteron = $(1.637+1.675)\times 10^{-27}=3.312\times 10^{-27}$

k = Boltzmann constant = $1.381\times 10^{-23}\ J/K$

T = Temperature = $2.94\times 10^8\ K$

RMS velocity is given by

$V_r=\sqrt{\dfrac{3kT}{m}}\\\Rightarrow V_r=\sqrt{\dfrac{3\times 1.381\times 10^{-23}\times 2.94\times 10^8}{3.312\times 10^{-27}}}\\\Rightarrow V_r=1917723.40119\ m/s$

The RMS velocity of the deutrons is 1917723.40119 m/s

8 0

2 years ago

Dark matter has been detected by its

Zinaida [17]

Dark matter has been detected by its gravitational pull.

7 0

3 years ago

A point charge q1 = -4.00 nC is at the point x = 0.60 m, y = 0.80 m , and a second point charge q2 = +6.00 nC is at the point x

makkiz [27]

Answer:

A)Magnitude of the net electric field at the origin due q₁ and q₂

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

β=3.91°

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

Data

q₁ = -4 nC = -4*10⁻⁹ C

q₂ = +6nC =+ 6*10⁻⁹ C

k = 9*10⁹ N*m²/C²

$d_{1} =\sqrt{0.6^{2}+0.8^{2} } =1 m$

d₂ = 0.6 m

Graphic attached

The attached graph shows the field in x=0, y=0, due to the charges q₁ and q₂:

E₁: Total field at point x=0 , y=0 due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge.

E₂: Total field at point x=0 , y=0 due to charge q₂. As the charge q2 is positive (q₂+) ,the field leaves the charge.

Calculation of the electric field at the origin of x-y coordinates due to the charge q₁

E₁=K*q₁/d₁²=9*10⁹*4*10⁻⁹/1²=36N/C

Components (x-y) of the field due to q1:

E₁x=E₁*cosα = 36*(0.6/1) = 36*(0.6) = 21.6 N/C

E₁y=E₁*sinα = 36*(0.8/1) = 36*(0.8) =28.8 N/C

Calculation of the electric field at the origin of x-y coordinates due to the charge q₂

E₂=K*q₂/d₂² = -9*10⁹*6*10⁻⁹/0.6² = -150 N/C

Calculation of the electric field components at the origin of x-y coordinates

E₀ is the electric field at the origin of x-y coordinates (x=0,y=0)

E₀x= E₁x+E₂= 21.6-150 = -128.4 N/C

E₀y= E₁y = 28.8 N/C

A )Magnitude of the net electric field at the origin due q₁ and q₂

$E_{o} =\sqrt{128.4^{2}+28.8^{2} }$

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

$\beta =tanx{-1} \frac{E_{oy} }{Eox}$

$\beta =tan^{-1} \frac{28.8}{128.4}$

β=3.91°

6 0

3 years ago