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1 year ago

5

Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration

due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

1 answer:

worty [1.4K]1 year ago

3 0

Answer:

3433.5 N

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of person = 70 kg

According to the question

a = Acceleration

$4g=4\times 9.81\\\Rightarrow a=39.24\ m/s^2$

Balancing the forces we have

$F-w=ma\\\Rightarrow F=ma+w\\\Rightarrow F=ma+mg\\\Rightarrow F=m(a+g)\\\Rightarrow F=70(39.24+9.81)\\\Rightarrow F=3433.5\ N$

The required force is 3433.5 N

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Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to

vesna_86 [32]

Answer:

The speed and direction of the apple is 1.448 m/s and 66.65°.

Explanation:

Given that,

Mass of apple = 0.110 kg

Speed = 1.13 m/s

Mass of orange = 0.150 kg

Speed = 1.25 m/s

Suppose we find the final speed and direction of the apple in this case

Using conservation of momentum:

Before:

In x direction,

$P_{b}=m_{p}v_{p}-m_{o}v_{o}$

$P_{b}=0.110\times1.13-0.150\times1.25$

$P_{b}=−0.0632\ kg-m/s$

In y direction = 0

After:

$v_{ay}$ is velocity of the apple in the y direction

$v_{ax}$ is the velocity of the apple in the x direction

Momentum again:

In x direction,

$0.110\times v_{ax}+0=−0.0632$

$v_{x}=\dfrac{−0.0632}{0.110}$

$v_{x}=−0.574\ m/s$

In y-direction,

$0.110\times v_{ay}-0.150\times0.977=0$

$v_{ay}=\dfrac{0.150\times0.977}{0.110}$

$v_{ay}=1.33\ m/s$

We need to calculate the speed of apple

$v_{a}=\sqrt{(v_{x})^2+(v_{y})^2}$

Put the value into the formula

$v_{a}=\sqrt{(−0.574)^2+(1.33)^2}$

$v_{a}=1.448\ m/s$

We need to calculate the direction of the apple

Using formula of angle

$\tan\theta=\dfrac{v_{ay}}{v_{ax}}$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{1.33}{0.574})$

$\theta=66.65^{\circ}$

Hence, The speed and direction of the apple is 1.448 m/s and 66.65°.

3 0

1 year ago

A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol

qaws [65]

Answer:

Part a)

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

$d = \frac{1}{2}gt^2$

Part c)

$v_f = v_o - gt$

Part d)

$d = \frac{1}{2}gt^2$

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

$v_y = -gt$

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

$d = \frac{1}{2}gt^2$

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

$v_f = v_i + at$

$v_f = v_o - gt$

Part d)

distance from helicopter

$d = \frac{1}{2}gt^2$

8 0

2 years ago

Read 2 more answers

You are on vacation in San Francisco and decide to take a cable car to see the city. A 5800-kgkg cable car goes 260 mm up a hill

Stella [2.4K]

Answer:

$4.325\times10^6J$

Explanation:

Mass of the cable car, m = 5800 kg

It goes 260 m up a hill, along a slope of $\theta=17^o$

Therefore vertical elevation of the car = $260sin\theta=260sin17^o=76.0166m$

Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = $\frac{1}{2} mv^2$ ). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.

Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.

Hence, total change in energy = mgh = $5800\times9.81\times76.0166J=4.325\times10^6J$

where, g = acceleration due to gravity

h = height/vertical elevation

4 0

1 year ago

The water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of

Damm [24]

Answer:

P_(pump) = 98,000 Pa

Explanation:

We are given;

h2 = 30m

h1 = 20m

Density; ρ = 1000 kg/m³

First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

Thus, it can be expressed as;

P_(tank)+ P_(pump) = P_(nozzle)

Now, the pressure would be given by;

P = ρgh

So,

ρgh_1 + P_(pump) = ρgh_2

Thus,

P_(pump) = ρg(h_2 - h_1)

Plugging in the relevant values to obtain;

P_(pump) = 1000•9.8(30 - 20)

P_(pump) = 98,000 Pa

5 0

1 year ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

2 years ago