What is this question

(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

2 years ago

A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

7 0

3 years ago

. A person weighing 750 N gets on an elevator.

Kobotan [32]

F = 750 N (Force)

d = 10 m (displacement )

t = 25 s (time)

L = ? (Mechanical work ) = (Energy)

P = ? (Power)

Solve:

L = F × d = 750 × 10 = 7500 Joules

P = L / t = 7500 / 25 = 300 Watts

5 0

2 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago

Which statement correctly describes the classification of chemical reactions into different categories?(1 point) Not all reactio

VMariaS [17]

Answer:

its a

Explanation:

4 0

3 years ago