Which best describes an acid?

(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us

konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

$R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m$

(b) The maximum height of the projectile in meters;

$H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m$

(c) The speed with which the projectile hits the ground is;

$v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s$

5 0

3 years ago

Often called velocity this is the velocity of an object at a particular moment in time

Yuliya22 [10]

Instantaneous speed?

8 0

3 years ago

Object 1 has a momentum of 10 kg m/s and Object 2 has a momentum of 25 kg m/s. Will it be easier to change the direction of move

svetoff [14.1K]

Answer:

I think its object 1

Explanation:

Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.

5 0

3 years ago

Which of the following is the best definition of energy?

Montano1993 [528]

Energy is the capacity to do some type of work

8 0

3 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

or

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

3 years ago