All categories

3 years ago

A student solving a physics problem for the range of a projectile has obtained the expression

Physics

1 answer:

Alika [10]3 years ago

6 0

Answer:

The range of the projectile is 66.7 meters.

Explanation:

The range of a projectile is given by the following expression as :

$R=\dfrac{v_o^2\ sin2\theta}{g}$ ..............(1)

$v_o=37.2\ m/s$

$\theta=14.1^{\circ}$

$g=9.8\ m/s^2$

The range can be calculated using equation (1). Putting the values of all parameters we get :

$R=\dfrac{(37.2)^2\ sin2(14.1)}{9.8}$

R = 66.7 meters

So, the range of the projectile is 66.7 meters. Hence, this is the required solution.

You might be interested in

Which force acts as an object to move it from rest or a constant straight line motion

sleet_krkn [62]

Answer:

The force is Inertia

Explanation:

The force that acts on an object to move it from rest or a constant straight line motion is known as Inertia.

In physics the above statement is governed by Newton's first law of motion which is also known as Law of Inertia.

This law states that, an object that is at rest will remain at rest and an object that is moving will continue to move in a straight line with constant speed, if and only if the net force acting on the object is zero.

This implies that, A stationary object will remain motionless if no force acts on it while a object with constant velocity will continue moving with constant velocity until a force acts on it (neglecting resistance from air and friction).

8 0

3 years ago

Read 2 more answers

Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6

Nana76 [90]

Answer:

The answer to the question is as follows

The acceleration due to gravity for low for orbit is 9.231 m/s²

Explanation:

The gravitational force is given as

$F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }$

Where $F_{G}$ = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹ $\frac{Nm^{2} }{kg^{2} }$

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by $F_{G}$ and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and $F_{G}$ = $\frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} }$ = 9.231 N

Therefore the acceleration due to gravity = $F_{G}$ /mass

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²

4 0

3 years ago

Read 2 more answers

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

4 0

3 years ago

Over a time interval of 1.99 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.7 km/

madam [21]

Answer:

(a) - 42700 m/s

(b) - 6.8 x 10^-4 m/s^2

Explanation:

initial velocity of star, u = 20.7 km/s

Final velocity of star, v = - 22 km/s

time, t = 1.99 years

Convert velocities into m/s and time into second

So, u = 20700 m / s

v = - 22000 m/s

t = 1.99 x 365.25 x 24 x 3600 = 62799624 second

(a) Change in planet's velocity = final velocity - initial velocity

= - 22000 - 20700 = - 42700 m/s

(b) Accelerate is defined as the rate of change of velocity.

Acceleration = change in velocity / time

= ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2

8 0

4 years ago

Conductors of large instrumental ensembles use thin stick called a __________ to help performers keep time.

Dmitry [639]

I think the word is "baton"

6 0

4 years ago

Read 2 more answers