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Y_Kistochka [10]

2 years ago

4

In an apparent effort to earn an appearance on the Destroyed in Seconds show, Caleb attempts a bicycle maneuver in which he jump

s between two ramps whose elevated edges are located a distance of 1.8 meters apart. The ramps are angled at 35° and located at the same height. Determine the speed (in m/s and mi/hr) that Caleb must acquire to accomplish this stunt. (Given: 1.00 m/s = 2.24 mi/hr)

1 answer:

Olenka [21]2 years ago

4 0

Answer:u=4.32 m/s

Explanation:

Given

Distance between two edges=1.8 m

launch angle $=35^{\circ}$

Let initial speed be u

Considering Caleb as projectile

$Range=\frac{u^2\sin 2\theta }{g}$

$u^2\sin 70=1.8\times 9.8$

$u^2=18.772$

$u=4.32 m/s=9.70 mi/hr$

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Suppose two astronauts on a spacewalk are floating motionless in space, 3.0 m apart. Astronaut B tosses a 15.0 kg IMAX camera to

marta [7]

Answer:

$\frac{ 112.5}{15+m_{A}}=v_{f}$

(we need the mass of the astronaut A)

Explanation:

We can solve this by using the conservation law of the linear momentum P. First we need to represent every mass as a particle. Also we can simplify this system of particles by considering only the astronaut A with an initial speed $v_{iA}$ of 0 m/s and a mass $m_{A}$ and the IMAX camera with an initial speed $v_{ic}$ of 7.5 m/s and a mass $m_{c}$ of 15.0 kg.

The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

$P_{i}=p_{ic}+p_{iA}\\P_{i}=m_{c}v_{ic}+m_{A} v_{iA}\\P_{i}=15*7.5 + m_{A}*0\\P_{i}=112.5 \frac{kg.m}{s}$

By the law of conservation we know that $P_{i} =P_{f}$

For $P_{f}$ (final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).

So:

$P_{i} =P_{f}=112.5\\$

$112.5=(m_{c}+m_{A})v_{f}\\\frac{ 112.5}{m_{c}+m_{A}}=v_{f}\\\frac{ 112.5}{15+m_{A}}=v_{f}$

3 0

2 years ago

The drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. A b

Advocard [28]

Answer:

Wet surfaces areaA=+25.3ft^2

Explanation:

Using F= K×A× S^2

Where F= drag force

A= surface area

S= speed

Given : F=996N S=20mph A= 83ft^2

K = F/AS^2=996/(83×20^2)

K= 996/33200 = 0.03

1215= (0.03)× A × 18^2

1215=9.7A

A=1215/9.7=125.3ft^2

7 0

2 years ago

Find the intensity in decibels [i(db)] for each value of i. normal conversation: i = 106i0 i(db) = power saw a 3 feet: i = 1011i

White raven [17]

Answer:

Normal Conversation: i=106i0

i(dB)=60

Power saw a 3 feet: i=1011i0

i(dB)=110

Jet engine at 100 feet: i=1018i0

i(dB)=180

Explanation:

if these are the same as edge, then these are the answers! :)

8 0

2 years ago

Read 2 more answers

A heat pump absorbs heat from the cold outdoors at 3Â°c and supplies heat to a house at 20Â°c at a rate of 30,000 kj/h. if the p

Mazyrski [523]

A heat pump absorbs heat from the cold outdoors at 3 C and supplies heat to a
house at 20 C at a rate of 30,000 kJ/h. If the power consumed by the heat pump
<span>is 3 kW, find the coefficient of performance of the heat pump.</span>

6 0

2 years ago

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

6 0

2 years ago