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2 years ago

15

Which statement best defines an electric field?

1 answer:

stich3 [128]2 years ago

3 0

Answer:

D

Explanation:

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The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps an

Marrrta [24]

Answer: $k=55.590 KN/m$

Explanation:

Given

mass of person $\left ( m\right )$ =68 kg

car dips about 1.2 cm

We know

F=kx

Where k=combined spring constant

mg=kx

$k=\frac{mg}{x}$

$k=\frac{68\times 9.81}{1.2\time 10^{-2}}$

$k=55.590 KN/m$

7 0

3 years ago

A tube of water is open on one end to the environment while the other end is closed. The height of the water relative to the bas

Andreas93 [3]

Answer:

<em>1.06 atm</em>

Explanation:

On the open end of the tube, the pressure will be the sum of atmospheric pressure and the pressure due to the height of water

The pressure due to a height of water = ρgh

where ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the height of the water column

The height of water column on the open end = 100 cm = 1 m

pressure on this end = ρgh = 1000 x 9.81 x 1 = 9810 Pa

Atmospheric pressure = 101325 Pa

The total pressure on the open end = 101325 Pa + 9810 Pa = <em>111135 Pa</em>

The pressure due to the water column on the closed end = ρgh

The height of the water in the closed end = 40 cm = 0.4 m

The pressure due to this column of water = 1000 x 9.81 x 0.4 = <em>3924 Pa</em>

The resultant pressure on the water on the top of the closed end of the tube = 111135 Pa - 3924 Pa =<em> 107211 Pa</em>

In atm unit, this pressure = 107211/101325 = <em>1.06 atm</em>

7 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:

<h3>b) Time</h3>

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

2 years ago

Will give correct answer brainliest

levacccp [35]

The potential energy= mass times gravity times height. However, we are missing height. Gravity is a constant that is 9.8 on Earth. We then solve for height by dividing 350 by 10 and then 9.8 to get about 3.5.
TLDR: 3.5

3 0

2 years ago

Read 2 more answers

A skier of mass 60 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 75 m up a 30° slop

postnew [5]

Answer:

Explanation:

Given

mass of skier=60 kg

distance traveled by skier=75 m

inclination $(\theta )=30^{\circ}$

speed (v)=2.4 m/s

as the skier is moving up with a constant velocity therefore net force is zero

$F_{net}=0$

Force applied by cable $=mg\sin \theta$

$F=60\times 9.8\times \sin (30)=294 N$

work done $=F\cdot x$

$W=294\cdot 75=22.125 J$

(b)Power $=F\cdot v$

$P=294\cdot 2.4=705.6 W\approx 0.946 hp$

5 0

2 years ago