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4 years ago

The equation below represents the process that you observed. Balance the equation. What type of reaction?

Physics

1 answer:

vladimir2022 [97]4 years ago

6 0

That looks like photosynthesis to me and I think that should be an endothermic reaction because it requires energy.

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Current is directly proportional to resistance.<br> a. True<br> b. False

ikadub [295]

IF voltage remains constant, then current is
inversely proportional to resistance.

The correct response is "b).", signifying "false" as the choice.

4 0

4 years ago

Pulsars are neutron stars that emit X rays and other radiation in such a way that we on Earth receive pulses of radiation from t

stira [4]

Answer:

(a) $a_{c} = 5.41\times 10^{9} m/s^{2}$

(b) $a_{t} = 2.99\times 10^{- 5} m/s^{2}$

Given:

Time period of Pulsar, $T_{P} = 33.085 ms == 33.085\times 10^{- 3} s$

Equatorial radius, R = 15 Km = 15000 m

Spinning time, $t_{s} = 9.50\times 10^{10}$

Solution:

(a) To calculate the value of the centripetal acceleration, $a_{c}$ on the surface of the equator, the force acting is given by the centripetal force:

$m\times a_{c} = \frac{mv_{c}^{2}}{R}$

$a_{c} = \frac{v_{c}^{2}}{R}$ (1)

where

$v_{c} = \frac{distance covered(i.e., circumference)}{ T}$

$v_{c} = \frac{2\pi R}{Time period, T}$ (2)

Now, from (1) and (2):

$a_{c} = R\frac({2\pi )^{2}}{T^{2}}$

$a_{c} = 15000\frac{2\pi )^{2}}{(33.085\times 10^{- 3})^{2}}$

$a_{c} = 5.41\times 10^{9} m/s^{2}$

(b) To calculate the tangential acceleration of the object :

The tangential acceleration of the object will remain constant and is given by the equation of motion as:

$v = u + a_{t}t_{s} = 0$

where

u = $v_{c}$

$a_{t} = - \frac{2\pi R}{Tt_{s}}$

$a_{t} = - \frac{2\pi 15000}{33.085\times 10^{- 3}\times 9.50\times 10^{10}}$

$a_{t} = 2.99\times 10^{- 5} m/s^{2}$

7 0

3 years ago

At what constant velocity must a spacecraft travel from earth if it is to reach the star in 3. 7 years, as measured by travelers

tatiyna

The constant velocity that the spacecraft must travel is : 3.49 * 10⁸ m/s

<u>Given data :</u>

Distance of star from earth = 4.3 light years

Observers time = 3.7 years

<h3>Determine the constant velocity the spacecraft must travel </h3>

Observers time = 3.7 * 365 * 24 * 60 * 60

Distance of star from earth = 4.3 * 9.46 * 10¹⁵

The velocity the spacecraft must travel will be calculated using the equation

V = distance / time

= ( 4.3 * 9.46 * 10¹⁵ ) / ( 3.7 * 365 * 24 * 60 * 60 )

= 3.49 * 10⁸ m/s

Hence we can conclude that The constant velocity that the spacecraft must travel is : 3.49 * 10⁸ m/s

Learn more about space travelling : brainly.com/question/1344685

<em />

<em>Attached below is the missing detail related to the question </em>

5 0

2 years ago

Read 2 more answers

Of the cloud types described here, which one cannot be easily classified as low, middle, or high?

diamong [38]

Cumulonimbus clouds

7 0

3 years ago

A truck with a mass of 1650 kg and moving with a speed of 15.0 m/s rear-ends a 779 kg car stopped at an intersection. The collis

ivanzaharov [21]

Answer:

Truck's speed = 5.21 m/s

Car's speed = 20.2 m/s

Explanation:

Given:

Mass of truck = M = 1650 kg

Speed of the truck initially = U = 15 m/s

Mass of the car = m = 779 kg

Initial speed of the car =u = 0

From the momentum conservation, Total initial momentum = Total final momentum.

M V+m U = M V +m v

⇒ (1650)(15) + 779×0 = (1650)V + 779 v

⇒ 24750 = 1650 V+779 v →(1)

Since the collision is elastic, relative velocity of approach = relative velocity of separation. 15 = v - V

⇒ v =V + 15; This is now substituted in the equation(1) above.

24750 = 1650 V + (799) (V+15)

⇒ 24750 = 1650 V + 799 V + 11985

⇒ 2449 V = 12765

⇒ Final velocity of the truck = $\frac{12765}{2449}$ = 5.21 m/s

Final velocity of the car = v = V+15 = 5.21 + 15 = 20.2 m/s

6 0

3 years ago