All categories

3 years ago

Explain a situation in which you can have a positive velocity but a negative acceleration

Physics

1 answer:

ziro4ka [17]3 years ago

7 0

A ball is thrown in the air. gravity is accelerating the ball downward while it still travels upwards

You might be interested in

What is an ionic bond?

Airida [17]

A.) a bond that forms when electrons are transferred from one atom to another

4 0

2 years ago

Read 2 more answers

g If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of th

Reil [10]

Answer:

Speed; v = 17 m/s

Explanation:

We are given;

Radius; r = 110m

Angle; θ = 15°

Now, we know that in circular motion,

v² = rg•tanθ

Thus,

v = √(rg•tanθ)

Where,

v is velocity

r is radius

g is acceleration due to gravity

θ is the angle

Thus,

v = √(rg•tanθ) = √(110 x 9.8•tan15)

v = √(288.85)

v = 17 m/s

8 0

3 years ago

4.A 31-cm long conducting wire of 9-g carrying 7- A current is placed in a uniform magnetic field. What are the strength and dir

Anika [276]

Answer

Given,

Length of the wire,L = 31 cm = 0.31 m

mass of the wire, m = 9 g = 0.009 Kg

Current in the wire,I = 7 A

Magnetic field strength, B= ?

Equating magnetic force to the weight of the wire.

$BIL = m g$

$B=\dfrac{m g}{IL}$

$B=\dfrac{0.009\times 9.81}{7\times 0.31}$

B = 0.0407 T

For Force to be upward magnetic field direction should be outward of the plane of paper.

7 0

3 years ago

A fair ground ride spins its occupants inside a flying saucer-shaped container. if the horizontal circular path the riders follo

Licemer1 [7]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below is the solution:

<span>centripetal accel = 1.5*g
ω²r = 1.5*9.8m/s²
ω² * 8m = 14.7 m/s²
ω = 1.36 rad/s * 1rev/2πrads * 60s/min = 12.9 rpm</span>

6 0

3 years ago

You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with

Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

Explanation:

For this exercise we use the constructor equation or Gaussian equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

m = $\frac{h'}{h}$ = - $\frac{q}{p}$

h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

$\frac{1}{q} = \frac{1}{f } - \frac{1}{p}$

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

h ’= h q / p

where p has a high value and is the same for all systems

h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

f = 12 cm h ’= fo 12 cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0

3 years ago