A ball is rolled from rest , find the velocity after 10s while accelerating at 5 m/s 2

What are AC and DC Power respectively used for?

kicyunya [14]

AC changes its direction in regular interval of time while DC is unidirectional flow. Due to many advantages of AC, it used to power our houses and offices while DC is used to power low power devices. AC is easier to transform between voltage levels, which makes high-voltage transmission more feasible

5 0

3 years ago

Oil of specific gravity 0.75 flows through a smooth contraction in a pipe at a volumetric flow rate of 3.2 cu ft / sec.Find the

pogonyaev

Answer:

The force required to hold the contraction in place is 665.91 N ↑

Explanation:

Given;

specific gravity of oil, γ = 0.75

Volumetric flow rate, V 3.2 Ft³/s = 0.0906 m³/s

$\gamma =\frac{\rho_o}{\rho_w}$

where;

$\rho_o$ is the density of oil

$\rho_w$ is the density of water = 1000 kg/m³

∴density of oil ( $\rho_o$ ) = γ × density of water( $\rho_w$ )

= 0.75 × 1000 kg/m³

= 750kg/m³

Buoyant Force = ρVg

= 750 × 0.0906 × 9.8

= 665.91 N ↑

This force acts upward or opposite gravitational force.

Therefore, the force required to hold the contraction in place is 665.91 N ↑

6 0

3 years ago

How are mitosis and meiosis similar?

fiasKO [112]

First year biology.

The answer is D. Each process results in the formation of Daughter Cells, except the process of Meiosis results in half the number of Daughter Cells, where Mitosis has Daughter Cells identical to the Parent Cell.

4 0

4 years ago

Read 2 more answers

A bicycle is traveling with a speed of 12 m/s. The cyclist hits the brakes reducing the speed to 4 m/s in a time of 4.8 seconds.

Annette [7]

So, the acceleration of the bicycle is approximately <u>-1.67 m/s²</u> or it can be said to be decelerating approximately <u>1.67 m/s²</u>.

<h3>Introduction</h3>

Hi ! Here I will help material about linear motion changes regularly, which is where you will hear a lot of the term acceleration. Acceleration occurs when an object's speed increases in a certain time interval. Acceleration can be negative which is called deceleration. The relationship between acceleration with velocity and time is manifested in the equation:

$\boxed{\sf{\bold{a = \frac{v_2 - v_1}{t}}}}$

With the following conditions :

a = acceleration (m/s²)
$\sf{v_2}$ = object's final velocity (m/s)
$\sf{v_1}$ = object's initial velocity (m/s)
t = interval of the time (s)

<h3>Problem Solving </h3>

We know that :

$\sf{v_2}$ = object's final velocity = 4 m/s
$\sf{v_1}$ = object's initial velocity = 12 m/s
t = interval of the time = 4.8 s

What was asked :

a = acceleration = ... m/s²

Step by step :

$\sf{a = \frac{v_2 - v_1}{t}}$

$\sf{a = \frac{4 - 12}{4.8}}$

$\sf{a = \frac{-8}{4.8}}$

$\boxed{\sf{a \approx -1.67 \: m/s^2}}$

So, the acceleration of the bicycle is about -1.67 m/s² or it can be said to be decelerating around 1.67 m/s².

8 0

3 years ago

An automobile battery has an emf of 12.6 V and an internal resistance of 0.0600 . The headlights together present equivalent res

murzikaleks [220]

Answer:

(a) $V=11.86\ V$

(b) $V=9.76\ V$

Explanation:

<u>Electric Circuits</u>

Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

$V=R.I$

(a) The electromagnetic force of the battery is $\varepsilon =12.6\ V$ and its internal resistance is $R_i=0.06\ \Omega$ . Knowing the equivalent resistance of the headlights is $R_e=5.2\ \Omega$ , we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

$\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)$

Solving for i

$\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A$

i=2.28\ A

The potential difference across the headlight bulbs is

$V=\varepsilon -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V$

$V=11.86\ V$

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

$V=\varepsilon -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V$

5 0

3 years ago