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3 years ago

Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly

Physics

1 answer:

loris [4]3 years ago

4 0

Answer: $36.86\°$

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: $d=2.4 m$

Ferdinand's initial velocity: $V_{o}=5 m/s$

Time it takes a jump: $t=0.6 s$

We need to find the angle $\theta$ at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

$V_{x}=\frac{d}{t}$ (1)

$V_{x}=\frac{2.4 m}{0.6 s}$ (2)

$V_{x}=4 m/s$ (3)

On the other hand, the x-component of the velocity is expressed as:

$V_{x}=V_{o}cos\theta$ (4)

Substituting (3) in (4):

$4 m/s=5 m/s cos\theta$ (5)

Clearing $\theta$ :

$\theta=cos^{-1} (\frac{4 m/s}{5 m/s})$

$\theta=36.86\°$ This is the angle at which Ferdinand the frog jumps between lily pads

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A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

3 years ago

1.How far is Object Z from the origin at t = 3 seconds?

GenaCL600 [577]

Answer:

Please find the answer in the explanation

Explanation:

1.) How far is Object Z from the origin at t = 3 seconds

The distance of the object Z from the origin will be the slope of the graph.

Slope = 4/2 = 2m

2.) Which object takes the least time to reach a position 4 meters from the origin ?

According to the graph given to the question above, object Z has the list time which is 2 seconds since object X does not start from the origin.

3.) Which object is farthest from the origin at t = 2 seconds?

The correct answer is still object Z because it has the highest slope.

4 0

3 years ago

Two circular coils are concentric and lie in the same plane. The inner coil contains 170 turns of wire, has a radius of 0.0095 m

PIT_PIT [208]

Answer:

Current in outer circle will be 15.826 A

Explanation:

We have given number of turns in inner coil $N_I=170$

Radius of inner circle $r_i=0.0095m$

Current in the inner circle $I_i=8.9A$

Number of turns in outer circle $N_o=150$

Radius of outer circle $r_o=0.015m$

We have to find the current in outer circle so that net magnetic field will zero

For net magnetic field current must be in opposite direction as in inner circle

We know that magnetic field is given due to circular coil is given by

$B=\frac{N\mu _0I}{2r}$

For net magnetic field zero

$\frac{N_I\mu _0I_I}{2r_I}=\frac{N_O\mu _0I_0}{2r_O}$

So $\frac{170\times \mu _0\times 8.9}{2\times 0.0095}=\frac{150\times \mu _0I_O}{2\times 0.015}$

$I_O=15.92A$

4 0

3 years ago

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t

Cloud [144]

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

$W = U_f - U_i$

we know that

$U_f = -\frac{GMm}{6r}$

$U_i = -\frac{GMm}{r}$

now from above formula

$W = -\frac{GMm}{6r} + \frac{GMm}{r}$

$W = \frac{5GMm}{6r}$

so above is the work done to move the mass from surface to given altitude

7 0

3 years ago

g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie

polet [3.4K]

Answer: $71.16\ Hz$

Explanation:

Given

Capacitance $C=100\ \mu F$

Resistance $R=500\ \Omega$

Inductance $L=50\ mH$

In LCR circuit, current is maximum at resonance frequency i.e.

$X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}$

Insert the values

$\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}$

Also, frequency is given by

$\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}$

$\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz$

8 0

3 years ago