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2 years ago

If an object has an initial velocity of 2.0 m/s and its rate of change in velocity is 3.0 m/s2

Physics

1 answer:

erica [24]2 years ago

6 0

Answer:

14m/s2

Explanation:

initial velocity=2m/s

final velocity =?

time=4s

acceleration=3.0m/s

v=u+at

v=(2)+(3*4)

v=14m/s

by substituting the values in the equation we get the value 14m/s.

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Suppose you have a Frisbee with a mass of 0.10 kg. You first throw it with a force of 5N and then change the force you throw it

natali 33 [55]

Answer:

The acceleration increases.

Explanation:

From Newton's 2nd Law, we have $\Sigma F=ma$ . We can see that force is directly proportional to mass and acceleration. Therefore, as force increases, either mass or acceleration must increase as well, and vice versa. Since mass is maintained here, if you increase the force applied to the Frisbee, the acceleration will increase as well.

8 0

3 years ago

Find the current in the 12 ohm resistor.

disa [49]

Answer:

1.5 A

Explanation:

V =I.R

18 = I × 12

I = 18/12

= 3/2 = 1.5 A

3 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago

The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres

12345 [234]

Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure $(P_1)$ =210+101.325=311.325

$T_1=25^{\circ}\approx 298 K$

$V=0.025 m^3$

$T_2=50^{\circ}\approx 323 K$

as the volume remains constant therefore

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{311.325}{298}=\frac{P_2}{323}$

$P_2=337.44 KPa$

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass $m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}$

$m_1=0.91 kg$

Final mass $m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}$

$m_2=0.839$

Therefore $m_1-m_2$ =0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

4 0

3 years ago

A 103 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 93 m below

siniylev [52]

Answer:

125.83672 seconds

Explanation:

P = Power of the horse = 1 hp = 746 W (as it is not given we have assumed the horse has the power of 1 hp)

m = Mass of professor = 103 kg

g = Acceleration due to gravity = 9.8 m/s²

h = Height of professor = 93 m

Work done would be equal to the potential energy

$W=mgh\\\Rightarrow W=103\times 9.8\times 93\\\Rightarrow W=93874.2\ J$

Power is given by

$P=\frac{W}{t}\\\Rightarrow t=\frac{W}{P}\\\Rightarrow t=\frac{93874.2}{746}\\\Rightarrow t=125.83672\ seconds$

The time taken by the horse to pull the professor is 125.83672 seconds

6 0

3 years ago