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Mamont248 [21]
3 years ago
8

Please answer giving all my points for it

Physics
2 answers:
Galina-37 [17]3 years ago
7 0

1. It would be 6

2. f - p (D)

The one on the bottom that you answered is right.

Papessa [141]3 years ago
6 0

A. 2

B. F - f

Hope this helps!

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Question 3 of 15
omeli [17]

Answer:

B) Degrees

Explanation:

The directions of the vectors are often defined in terms of due East, due North, due West and due South. A direction exactly in between of North and East can be described as Northeast, similarly we can describe directions in terms of Northwest, Southeast and South west.

From these, the direction of a vector can be easily expressed in degrees, which is measured counter clockwise about its tail from due East. Considering that we can say that East is at 0° , North is at 90° , West is at 180 and South is at 270° counter clockwise rotation from due East.

So, we know that the direction of a vector lying somewhere between due East i.e 0° and due North i.e 90°, will be measured in degrees, which will have a value between 0°-90°

4 0
2 years ago
Which shows the formula for converting from kelvins to degrees Celsius? °C = (9/5 × K) +32 °C = 5/9 × (K – 32) °C = K – 273 °C =
madam [21]

The freezing point of the water is 0 C , and it equals to 273 K

Then, To convert from Kelvins degrees to Celsius degrees we use the relation

K = C + 273

Also,

C = K - 273

5 0
3 years ago
Read 2 more answers
Hi, I need some science help ASAP. Here's the question:
ch4aika [34]

Answer:

Mass of the box is 25kg

Explanation:

From Newton's second law of motion

Force= mass. acceleration

With the symbol

F=m.a

m=F/a      m=(50N)/(2m/s²)

                m=25kg

7 0
3 years ago
Which one of these is an exanple of precepation?
Leokris [45]

Answer:

Snow

Explanation:

Precipitation is the formation of a solid after being a liquid. Snow, which is a solid, forms from water, a liquid.

8 0
2 years ago
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Elect the correct answer. Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of laun
allsm [11]

Answer:

A. The range of A and B are equal.

Explanation:

Let u be the initial speed of both the projectile.

The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.

For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,

The speed in the horizontal direction = u\cos\theta

and the speed in the vertical direction is = u\sin\theta upward.

For A:

The speed in the horizontal direction = u\cos75^{\circ}

and the speed in the vertical direction is = u\sin75^{\circ} upward.

For B:

The speed in the horizontal direction = u\cos15^{\circ}

and the speed in the vertical direction is = u\sin15^{\circ} upward.

Let t_A and t_B are the time of flight for projectile A and B respectively.

As the range is the horizontal distance traveled by the projectile, so

The range for the projectile A = u\cos75^{\circ}\times t_A\cdots(i)

The range for the projectile B = u\cos15^{\circ}\times t_B\cdots(ii)

At the highest point, the vertical velocity is 0.

Bu using the equation of motion v^2=u^2 +2a s.

Here, the final velocity v=0, the initial velocity u = u \sin \theta , h= vertical distance up to the highest point, and a= -g (as per sign convention).

So, s= \frac{u^2\sin^2 \theta}{2g}

For projectile A: The maximum height attained.

s_A= \frac{u^2\sin^2 75^{\circ}}{2g}

For projectile B: The maximum height attained.

s_B= \frac{u^2\sin^2 15^{\circ}}{2g}

As \sin^2 75^{\circ} > \sin^2 15^{\circ}, the height of A is attained by A is more than the heigHt attained by B.

Now, the times required to reach the highest point from the ground and again form the highest point to the ground are the same.

So, the total time of flight = 2 x (Time to reach the highest point)

In a similar way, by using the equation of motion v=u+at,

The time to reach the highest point =\frac {u\sin\theta}{g}

where g is the acceleration due to gravity.

So, the total time of flight

= 2 \times \frac {u\sin\theta}{g}

The total time of flight for A

=2 \times \frac {u\sin75^{\circ}}{g}

The total time of flight for A

=2 \times \frac {u\sin15^{\circ}}{g}

Now, from equations (i) and (ii),

The range for the projectile A =

u\cos75^{\circ}\times  \frac {2u\sin75^{\circ}}{g}=\frac {u^2 \sin 150^{\circ}}{g}= \frac {u^2 \sin 30^{\circ}}{g}

The range for the projectile B =

u\cos15^{\circ}\times  \frac {2u\sin15^{\circ}}{g}=\frac {u^2 \sin 30^{\circ}}{g}.

Both the projectile have the same range.

Hence, option (A) is correct.

7 0
3 years ago
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