The weigh of any object is computed by multiplying its mass to the acceleration of gravity, so we need to find the gravity on that planet in order to compute the weigh we want.

The ball has a mass of 0.1 kg and its released from a height of 10 m, therefore it is in a free fall motion with gravity acting as a constant acceleration on the body, we can use the equations for free fall movement in order to determine the value for this acceleration:

y(t) = v_0 * t + y_0 - 0.5 * g * t^2

y(t) is the position in the end of the movement, when t = 3.4 s, so y(t) = 0 m.

v_0 is the initial velocity, in this case v_0 = 0 m/s.

y_0 is the initial position of the ball, in this case it is 10 m.

g is the gravity that we want to know.

Applying these values in the equation we have:

0 = 0*(3.4) + 10 - 0.5*g*(3.4)^2

0 = 10 - 0.5*11.56*g

0 = 10 -5.78*g

5.78*g = 10

g = 1.73 m/s^2

Then we can use this value to find out the weigh of the ball in that planet:

w = g*m = 0.1*1.73 = 0.173 N

6 0

4 years ago

I will gib brainlyest or whatever.

astraxan [27]

Answer:

Range of the projectile: approximately $1.06 \times 10^{3}\; {\rm m}$ .

Maximum height of the projectile: approximately $80\; {\rm m}$ (approximately $45.0\; {\rm m}$ above the top of the cliff.)

The projectile was in the air for approximately $7.07\; {\rm s}$ .

The speed of the projectile would be approximately $155\; {\rm m \cdot s^{-1}}$ right before landing.

(Assumptions: drag is negligible, and that $g = 9.81\; {\rm m\cdot s^{-1}}$ .)

Explanation:

If drag is negligible, the vertical acceleration of this projectile will be constantly $a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}$ . The SUVAT equations will apply.

Let $\theta$ denote the initial angle of elevation of this projectile.

Initial velocity of the projectile:

vertical component: $u_{y} = u\, \sin(\theta) = 153\, \sin(11.2^{\circ}) \approx 29.71786\; {\rm m\cdot s^{-1}}$
horizontal component: $u_{x} = u\, \cos(\theta) = 153\, \cos(11.2^{\circ}) \approx 150.086\; {\rm m\cdot s^{-1}}$ .

Final vertical displacement of the projectile: $x_{y} = (-35)\; {\rm m}$ (the projectile landed $35\: {\rm m}$ below the top of the cliff.)

Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the final vertical velocity $v_{y}$ of this projectile:

${v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}$ .

$\begin{aligned} v_{y} &= -\sqrt{{u_{y}}^{2} + 2\, a_{y} \, x_{y}} \\ &= -\sqrt{(29.71786)^{2} + 2\, (-9.81)\, (-35)} \\ &\approx (-39.621)\; {\rm m\cdot s^{-1}}\end{aligned}$ .

(Negative since the projectile will be travelling downward towards the ground.)

Since drag is negligible, the horizontal velocity of this projectile will be a constant value. Thus, the final horizontal velocity of this projectile will be equal to the initial horizontal velocity: $v_{x} = u_{x}$ .

The overall final velocity of this projectile will be:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \sqrt{(150.086)^{2} + (-39.621)^{2}} \\ &\approx 155\; {\rm m\cdot s^{-1}} \end{aligned}$ .

Change in the vertical component of the velocity of this projectile:

$\begin{aligned} \Delta v_{y} &= v_{y} - u_{y} \\ &\approx (-39.621) - 29.71786 \\ &\approx 69.3386 \end{aligned}$ .

Divide the change in velocity by acceleration (rate of change in velocity) to find the time required to achieve such change:

$\begin{aligned}t &= \frac{\Delta v_{y}}{a_{y}} \\ &\approx \frac{69.3386}{(-9.81)} \\ &\approx 7.0682\; {\rm s}\end{aligned}$ .

Hence, the projectile would be in the air for approximately $7.07\; {\rm s}$ .

Also the horizontal velocity of this projectile is $u_{x} \approx 150.086\; {\rm m\cdot s^{-1}}$ throughout the flight, the range of this projectile will be:

$\begin{aligned}x_{x} &= u_{x}\, t \\ &\approx (150.086)\, (7.0682) \\ &\approx 1.06 \times 10^{3}\; {\rm m} \end{aligned}$ .

When this projectile is at maximum height, its vertical velocity will be $0$ . Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the maximum height of the projectile (relative to the top of the $35\; {\rm m}$ cliff.)

$\begin{aligned}x &= \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a} \\ &\approx \frac{0^{2} - 29.71786^{2}}{2\, (-9.81)} \\ &\approx 45.0\; {\rm m}\end{aligned}$ .

Thus, the maximum height of the projectile relative to the ground will be approximately $45.0\; {\rm m} + 35\; {\rm m} = 80\; {\rm m}$ .

5 0

1 year ago

7. A 78-kg skydiver has a speed of 62 m/s at an altitude of 870 m above the ground.

miv72 [106K]

The kinetic energy, potential energy and total mechanical energy possessed by the skydriver are 1.5 × 10⁵ Joule, 6.7 × 10⁵ Joule and 8.2 × 10⁵ Joule respectively.

To find the answer, we need to know about the expression of kinetic energy and potential energy.

<h3>What are the expressions of kinetic energy and potential energy?</h3>

Mathematically, kinetic energy= 1/2 × mass × velocity²
Gravitational potential energy near the earth surface= mass × g × height on the earth surface

<h3>What's the kinetic energy, potential energy and total mechanical energy of the 78kg skydriver at 870 m on earth surface with 62 m/s velocity?</h3>

Kinetic energy= 1/2 × 78 × 62² = 1.5 × 10⁵ Joule
Potential energy= 78×9.8×870= 6.7× 10⁵ Joule

<h3>What's the total mechanical energy?</h3>

Mechanical energy= kinetic energy+ potential energy
1.5 × 10⁵ Joule + 6.7× 10⁵ Joule = 8.2× 10⁵ Joule

Thus, we can conclude that the kinetic energy, potential energy and total mechanical energy possessed by the skydriver are 1.5 × 10⁵ Joule, 6.7 × 10⁵ Joule and 8.2 × 10⁵ Joule respectively.

Learn more about the kinetic energy, potential energy and mechanical energy here:

brainly.com/question/17051553

#SPJ1

8 0

2 years ago

Help me frfr I don’t understand

Art [367]

The points are

(1,10)

(6,0)

$\boxed{\sf slope(m)=\dfrac{y_2-y_1}{x_2-x_1}}$