Which statements about acceleration are true?

The newest generation of smart phone uses three different batteries connected in a series simultaneously to extend battery life

azamat

Answer:

7.2 V

Explanation:

The three batteries are connected in series to the terminals of the phone: it means that they are connected along the same branch, so the current flowing through them is the same.

This also means that the potential difference across the phone will be equal to the sum of the voltages provided by each battery.

Here, the voltage provided by each battery is

V = 2.4 V

So, the overall voltage will be

V = 2.4 V + 2.4 V + 2.4 V = 7.2 V

8 0

3 years ago

Most earthquakes occur in areas close to where tectonic plates meet. There are earthquakes in San Francisco. What can be conclud

earnstyle [38]

B. To have an earthquake there must be a fault line (where two or more tectonic plates meet) so if San Fran. has earthquakes they’re on a fault line.

5 0

3 years ago

Read 2 more answers

A baseball is batted from a height of 1.09 m with a speed of

kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

$H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m$

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

$V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s$

<h3>Resultant speed of the ball and crosswind</h3>

$V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s$

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0

2 years ago

A 2,000 kg car starts from rest and coasts down from the top of a 5.00 m long driveway that is sloped at an angel of 20o with th

ehidna [41]

Answer:

The speed at the bottom of the driveway is3.67m/s.

Explanation:

Height,h= 5sin20°= 1.71m

Potential energy PE=mgh= 2000×9.8×1.71

PE= 33516J

KE= PE- Fk ×d

0.5mv^2= 33516 - (4000×5)

0.5×2000v^2= 33516 - 20000

1000v^2= 13516

v^2= 13516/1000

v =sqrt 13.516

v =3.67m/s

5 0

3 years ago

An astronaut on a space walk floats a little too far away from the space station. Without air to push against, he cannot paddle

Artyom0805 [142]

He can throw the hammer in the direction opposite to the direction he wants to travel in. The hammer will exert an equal and opposite force on him, as per Newton's third law, and this will help him move towards the space station.

5 0

3 years ago

Read 2 more answers