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3 years ago

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PLEASE HELP An automobile with an initial speed of 5.80 m/s accelerates uniformly at the rate of 4.1 m/s^2 Find the final speed

of the car after 5.6 s. Answer in units of m/s. Find the displacement of the car after 5.6 s. Answer in units of m.

2 answers:

Lemur [1.5K]3 years ago

8 0

Answer:

dfuvadfivu

Explanation:

e-lub [12.9K]3 years ago

6 0

Thanks for asking the question

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cupoosta [38]

Answer:

<h3>The answer is 1.92 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 2.5 g

Volume = 1.3 cm³

We have

$density = \frac{2.5}{1.3} \\ = 1.923076...$

We have the final answer as

<h3>1.92 g/cm³</h3>

Hope this helps you

5 0

3 years ago

Please answer the questions here in the attachment I attached below. They are all multiple choice.

Radda [10]

Answer:

abcbc

Explanation:

8 0

4 years ago

A projectile is fired at an upward angle of 29.7Â° from the top of a 108-m-high cliff with a speed of 130-m/s. What will be its

german

Answer:

79.2 m/s

Explanation:

θ = angle at which projectile is launched = 29.7 deg

a = initial speed of launch = 130 m/s

Consider the motion along the vertical direction

v₀ = initial velocity along the vertical direction = a Sinθ = 130 Sin29.7 = 64.4 m/s

y = vertical displacement = - 108 m

a = acceleration = - 9.8 m/s²

v = final speed as it strikes the ground

Using the kinematics equation

v² = v₀² + 2 a y

v² = 64.4² + 2 (-9.8) (-108)

v = 79.2 m/s

5 0

3 years ago

A particle of mass m = 13 kg moves in space under the action of a conservative force. Its potential energy is given by PE = 2xyz

Helen [10]

Answer:

$F_{x}$ = -12 N , $F_{y}$ = -80 N and $F_{z}$ = - 44 N

Explanation:

The force is related to the potential energy by the formula

F = -Δ U = - ( $\frac{dU}{dx}$ i ^ + $\frac{dU}{dy}$ j ^ + $\frac{dU}{dz}$ k ^)

It indicates the potential energy

U = 2xyz + 3z² + 4yx + 16

To solve this problem let's make the derivatives, to find each component of the force

$\frac{dU}{dx}$ = 2yz + 0 + 4y + 0 = 2yz + 4y

$\frac{dU}{dx}$ = 2xz + 0 + 4x + 0 = 2xz + 4x

$\frac{dU}{dx}$ = 2xy + 3 2z + 0 +0 = 2xy + 6z

We look for the expression for the force in each axis

$F_{x}$ = - 2yz - 4y

$F_{y}$ = -2xz -4x

$F_{z}$ = -2xy -6z

We calculate at the point P = (20 i ^ + 1j ^ + 4 k ^) m

$F_{x}$ = - 2 1 4 - 4 1

$F_{x}$ = -12 N

$F_{y}$ = - 2 20 4 - 4 20

$F_{y}$ = -80 N

$F_{z}$ = - 20 1 - 6 4

$F_{z}$ = - 44 N

We put together the expression for strength

F = (-12 i ^ - 80j ^ -44k ^) N

5 0

4 years ago

Eld a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1

zheka24 [161]

Answer:

$v_{f}=1721.1m/s$

Explanation:

Given data

q = 3.1 µC

m = 47 mg

r1 = 0.83 mm

r2 = 2.5 mm.

As we know that:

$dK=-dU\\(1/2)mv_{f}^{2}-(1/2)mv_{i}^{2}=-(\frac{kq^{2} }{r_{f}}-\frac{kq^{2} }{r_{i}} )\\ (1/2)*(47*10^{-6}kg )v_{f}^{2}-(1/2)*(47*10^{-6}kg)(0)=-[(\frac{(9*10^{9}(3.1*10^{-6})^{2} }{2.5*10^{-3}}-(\frac{(9*10^{9}(3.1*10^{-6})^{2} }{0.83*10^{-3}} )]\\2.35*10^{-5} v_{f}^{2}=69.61\\v_{f}=\sqrt{\frac{69.61}{2.35*10^{-5}} }\\v_{f}=1721.1m/s$

5 0

3 years ago