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3 years ago

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interval (from lowering the body to his feet taking off) lasts for 0.3 s and the mass of the player is 90 kg. Ignore air resista

nce. Part A Determine the force that the the player exerts on the surface when he jumps. Express your answer to one significant figure and include the appropriate units. Enter positive value if the force is upward and negative value if the force is downward.

1 answer:

steposvetlana [31]3 years ago

8 0

Answer:

The force applied to the surface is 9 kilo Newton.

Explanation:

While jumping on the surface the player applies the force that is equal to its weight on the surface.

The mass of the player is given as 90 kg.

Force applied by the player = weight of the player

Force applied by the player = m × g

Where m is the mass of the player and g is acceleration due to gravity

Force applied by the player = 90 × 9.8

Force applied by the player = 882 Newton

Expressing your answer to one significant figure, we get

Force applied by the player =0. 9 kilo Newton

The force applied to the surface is 0.9 kilo Newton.

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An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

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3 years ago

Dana is on a train traveling at a speed of 20 km/h. Dana walks from the front of the train to the back of the train at a speed o

Maslowich

Answer:

16km/h

Explanation:

Vt=20km/h ---train speed

Vd=4km/h

Donas speed relative to ground is:

Vrd=Vt-Vd

Donas is moving in opposite direction of train .

Vrd=20km/h-4km/h

Vrd=16km/h

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4 years ago

Chris threw a basketball a distance of 27.5 m to score and win his

salantis [7]

Answer:

v₀ = 16.55 m/s

Explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

θ = Launch Angle = 50°

g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

$R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}$

<u>v₀ = 16.55 m/s</u>

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3 years ago

The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the part

svetoff [14.1K]

Answer:

B. distance/potential

Explanation:

Quizlet

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3 years ago

Consider the following three objects, each of the same mass and radius: (1) a solid sphere (2) a solid disk (3) a hoop All three

Vinil7 [7]

Answer:

The correct answer is

a) 1, 2, 3

Explanation:

In rolling down an inclined plane, the potential energy is Transferred to both linear and rotational kinetic energy thus

PE = KE or mgh = 1/2×m×v² + 1/2×I×ω²

The transformation equation fom potential to kinetic energy is =

m×g×h = $\frac{1}{2} mv^{2} + \frac{1}{2} (\frac{2}{5} mr^{2} )(\frac{v}{r}) ^{2}$

$v_{Sphere}$ = $\sqrt{\frac{10}{7} gh}$

$v_{Hoop}$ = $\sqrt{gh}$

$v_{Disc}$ = $\sqrt{\frac{4}{3} gh}$

Therefore the order is with increasing rotational kinetic energy hence

the first is the sphere 1 followed by the disc 2 then the hoop 3

the correct order is a, 1, 2, 3

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4 years ago