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3 years ago

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A car driving on the turnpike accelerates uniformly in the positive x direction from 80 ft/s (54.5 mph) to 103 ft/s (70.2 mph) i

n 3.10s. What is the x- component of the car's acceleration ?

1 answer:

Oxana [17]3 years ago

4 0

Acceleration = (change in speed) / (time for the change)

Change in speed = (ending speed) - (starting speed)

= (103 fps) - (80 fps) = 23 ft/sec .

Acceleration = (23 ft/sec) / (3.1 sec) = 7.42 ft/sec² .

Since the acceleration was uniform in the x-direction,
its component in the x-direction is 7.42 ft/sec², and it
has no other component.

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A sphere of radius 0.081029 m is made of aluminum. It is submerged in flowing water with a temperature of 25 °C that results in

Dimas [21]

Answer:

its surface temperature = 54.84 ° C

Explanation:

The density of aluminium $(\rho)$ = 2700 kg/m ³

Heat capacity $( c_p)$ = 897 J/Kg.K

radius of the sphere (r) = 0.081029 m

$T \infty$ = 25 °C

$T_i$ = 124.978 °C

time (t) = 767.276 s

Using the formula :

$\frac{T-T_{ \infty} }{T_i - T_{\infty}}= e^{-\frac{hA}{\rho V c_p}}*t$

where.

$\frac{V}{A}= \frac{r}{3}$

Replacing our values ;we have:

$\frac{T-25 }{124.978 - 25}= e^{-\frac{-103.067*3}{2700*897*0.081029}}*767.276$

$\frac{T-25 }{124.978 - 25}= e^{{-0.001576}*767.276$

$\frac{T-25 }{124.978 - 25}= e^{-1.209}$

$\frac{T-25 }{99.978}= 0.2985$

${T-25 }= 0.2985*{99.978}$

${T-25 }= 29.843433$

${T= 29.843433+25 }$

${T= 54.843433$

T ≅ 54.84 ° C

Therefore, its surface temperature = 54.84 ° C

6 0

4 years ago

A step-down transformer has more loops on which coil?

Nesterboy [21]

A step-down transformer has more loops in : A. Primary coil

Primary coil refers to the coil to which alternating voltage is supplied. It's usually connected to the AC supply

hope this helps

6 0

4 years ago

Read 2 more answers

When a small object is launched from the surface of a fictitious planet with a speed of 52.9 m/s, its final speed when it is ver

NISA [10]

Answer:

The escape speed of the planet is 41.29 m/s.

Explanation:

Given that,

Speed = 52.9 m/s

Final speed = 32.3 m/s

We need to calculate the launched with excess kinetic energy

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

$K.E=\dfrac{1}{2}\times m\times(32.3)^2$

We need to calculate the escape speed of the planet

Using formula of kinetic energy

$\text{escape kinetic energy}=\text{launch kinetic energy}-\text{excess kinetic energy}$

$\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2-\dfrac{1}{2}mv^2$

$\dfrac{1}{2}\times v^2=\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2$

$v=\sqrt{2\times(\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2)}$

$v=41.29\ m/s$

Hence, The escape speed of the planet is 41.29 m/s.

4 0

3 years ago

Physics question, please answer this ASAP, thanks!

miv72 [106K]

(a) The gravitational force received by each 1 kg mass is 8.66 N.

(b) The magnitude of gravitational acceleration is 8.66 m/s².

(c) The orbital speed of the ISS is 7,663.6 m/s.

(d) The time take for the ISS to orbit round the Earth is 5,558.75 s = 1.54 hours.

<h3>Gravitational force received by each 1 kg mass</h3>

The gravitational force received by each 1 kg mass is calculated as follows;

F = Gm₁m₂/r²

where;

m₁ is mass of Earth
m₂ is mass of ISS
r is the distance between the ISS and center of Earth

F = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ x 1) / (6780,000)²

F = 8.66 N

<h3>Magnitude of gravitational acceleration</h3>

mg = GMm/r²

g = GM/r²

where;

M is mass of Earth

g = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ ) / (6780,000)²

g = 8.66 m/s²

<h3>Orbital Speed of the ISS</h3>

v = √GM/r

v = √(6.67 x 10⁻¹¹ x 5.97 x 10²⁴ / 6780,000)

v = 7,663.6 m/s

<h3>Time of motion of the ISS round the Earth</h3>

T = 2πr/v

T = (2π x 6780,000) / (7663.6)

T = 5,558.75 seconds

1 hour = 3600 seconds

= 5,558.75/3600

= 1.54 hours

Thus, the gravitational force received by each 1 kg mass is 8.66 N.

The magnitude of gravitational acceleration is 8.66 m/s².

The orbital speed of the ISS is 7,663.6 m/s.

The time take for the ISS to orbit round the Earth is 5,558.75 s = 1.54 hours.

Learn more about orbital speed here: brainly.com/question/22247460

#SPJ1

4 0

2 years ago

Points P and Q are located at (0, 2, 4) and (-3, 1,5). Calculate

Akimi4 [234]

Answer:

a) The position vector of P is $\vec P =(0, 2,4)$ .

b) The distance vector from P to Q is $\overrightarrow{PQ} = (-3,-1,1)$ .

c) The distance between P and Q is $\|\overrightarrow{PQ}\|=\sqrt{11}$ .

d) A vector parallel to PQ with magnitude of 10 is $\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$ .

Explanation:

a) The position vector of a point is the vector displacement from the origin to the location of the point. That is:

$\vec P = (0,2,4)-(0,0,0)$

$\vec P = (0-0, 2-0, 4-0)$

$\vec P =(0, 2,4)$

The position vector of P is $\vec P =(0, 2,4)$ .

b) First, we calculate the position vector of point Q:

$\vec Q = (-3,1,5)-(0,0,0)$

$\vec Q = (-3-0,1-0,5-0)$

$\vec Q =(-3,1,5)$

The distance vector from P to Q is define by the following vectorial expression:

$\overrightarrow{PQ} = \vec Q - \vec P$ (1)

$\overrightarrow{PQ} = (-3,1,5)-(0,2,4)$

$\overrightarrow{PQ} =(-3-0,1-2,5-4)$

$\overrightarrow{PQ} = (-3,-1,1)$

The distance vector from P to Q is $\overrightarrow{PQ} = (-3,-1,1)$ .

c) There are two approaches to calculate the distance between P and Q:

First Method - Pythagorean Theorem:

$\|\overrightarrow{PQ}\| = \sqrt{(-3)^{2}+(-1)^{2}+1^{2}}$

$\|\overrightarrow{PQ}\|=\sqrt{11}$

Second Method - Dot Product:

$\|\overrightarrow{PQ}\| = \sqrt{\overrightarrow{PQ}\,\bullet\,\overrightarrow{PQ}}$ (2)

$\|\overrightarrow{PQ}\| = \sqrt{(-3,-1,1)\,\bullet (-3,-1,1)}$

$\|\overrightarrow{PQ}\|=\sqrt{11}$

The distance between P and Q is $\|\overrightarrow{PQ}\|=\sqrt{11}$ .

d) To determine a vector parallel to PQ with a given magnitude is determined by the following expression:

$\vec v = \frac{k}{\|\overrightarrow{PQ}\|} \cdot \overrightarrow{PQ}$ (3)

Where $k$ is the scale factor.

If we know that $\overrightarrow{PQ} = (-3,-1,1)$ , $\|\overrightarrow{PQ}\|=\sqrt{11}$ and $k = 10$ , then the vector is:

$\vec v = \frac{10}{\sqrt{11}}\cdot (-3,-1,1)$

$\vec v = \left(-\frac{30}{\sqrt{11}},-\frac{10}{\sqrt{11}},\frac{10}{\sqrt{11}}\right)$

$\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$

A vector parallel to PQ with magnitude of 10 is $\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$ .

8 0

3 years ago