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dimaraw [331]
3 years ago
6

Katie, a 40 kg child, climbs a tree to rescue her cat who is afraid to jump 9.4 m to the ground. How much work does Katie do to

climb the tree in order to reach the cat? Answer to the nearest Joule.
Physics
1 answer:
Diano4ka-milaya [45]3 years ago
4 0

Answer:

The work done by the child to reach the cat is, W = 3684.8 J        

Explanation:

Given data,

The mass of the child, m = 40 kg

The height of the cat from the ground, h = 9.4 m

The force acting on the child due to gravitational force is,

                                  F = mg

The minimum force required by the child to climb the tree against the gravitational force is,

                                 F = mg

Hence, the work done by the child is,

                                W = F ·  h

                                     = 40 x 9.8 x 9.4

                                     = 3684.8 joules

Hence, the work done by the child to reach the cat is, W = 3684.8 J                      

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A force of 14 N acts on a 5 kg object for 3 seconds.
DiKsa [7]

Answer: a) 42Nm b) 8.4m/s

Explanation:

Impulse is defined as object change in momentum.

Since Force = mass × acceleration

F = ma

Acceleration is the rate of change in velocity.

F = m(v-u)/t

Cross multiply

Ft = m(v-u)

Since impulse = Ft

and Ft = m(v-u)... (1)

The object change in velocity (v-u) = Ft/m from eqn 1

Going to the question;

a) Impulse = Force (F) × time(t)

Given force = 14N and time = 3seconds

Impulse = 14×3

Impulse = 42Nm

b) The object change in velocity (v-u) = Ft/m where mass = 5kg

v-u = 14×3/5

Change in velocity = 42/5 = 8.4m/s

3 0
3 years ago
Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano
Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

<h2>Part a </h2>

Specific energy is given by

\epsilon=\frac{v^2}{2}-\frac{\mu}{r}

Here

  • ε is the specific energy
  • v is the velocity which is given as 2.23 km/s
  • μ is the gravitational constant whose value is 398600
  • r is the distance between earth and the meteorite which is 402,000 km

                         \epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2

Value of specific energy is also given as

\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km

Orbit formula is given as

r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0

Putting values in this equation and solving for e via the quadratic formula gives

ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

<h2>Part b</h2>

The radius of trajectory at perigee is given as

r_p=a(e-1)\\

Substituting values gives

r_p=133319 (1.086-1)\\r_p=11465.4 km

Now for estimation of altitude z above earth is given as

z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km

So the altitude at closest approach is 5088 km

<h2>Part c</h2>

radius of perigee is also given as

r_p=\frac{h^2}{\mu}\frac{1}{1+e}

Rearranging this equation gives

h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s

Now the velocity at perigee is given as

v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\

So the velocity at perigee is 8.516 km/s

<h2>Part d</h2>

Turn angle is given as

\delta =2 sin^{-1} (\frac{1}{e})

Substituting value in the equation gives

\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08

Aiming radius is given as

\Delta =a \sqrt{e^2-1}

Substituting value in the equation gives

\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0
4 years ago
When current first leaves the battery, how much voltage does it have
rjkz [21]

Answer:

depends on the voltage of battery

Explanation:

for example if you connect a battery of 6V,6V will be  provided

8 0
3 years ago
A torch is dropped down a large chasm. It takes 10 seconds to hit the ground. How deep is the chasm?
ale4655 [162]

Answer: 490m

Explanation: 1/2 * 9.8m/s/s * 10s

3 0
4 years ago
If 400g is 1kg find the ratio in the simplest form​
uranmaximum [27]

2:5

Explanation:

400g : 1kg

400g: 1000g

4 : 10

2 : 5

6 0
3 years ago
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