Which nucleus completes the following equation?

A boat heads directly across a river. Its speed relative to the water is 2.6 m/s. It takes it 355 seconds to cross, but it ends

Zigmanuir [339]

Answer:

Vr = 3.24m/s

The boat is going 3.24m/s relative to the bank of the river.

Explanation:

The relative speed of the boat to the bank Vr is the resultant of speed of boat relative to the water Vb and the speed of boat as a result of the water current or wind Vw

Vr = √(Vb^2 + Vw^2) .....1

Given;

Vb = 2.6m/s

Vw = distance downstream/time = 690m/355s

Vw = 1.94m/s

From equation 1 above; substituting the values

Vr = √(2.6^2 + 1.94^2)

Vr = 3.24m/s

The boat is going 3.24m/s relative to the bank of the river.

7 0

3 years ago

g The “size” of the atom in Rutherford’s model is about 8 × 10−11 m. Determine the attractive electrostatics force between a ele

Tju [1.3M]

Answer:

$3.6\cdot 10^{-8} N$

Explanation:

The electrostatic force between the proton and the electron is given by:

$F=k\frac{q_p q_e}{r^2}$

where

$k=9.00\cdot 10^9 Nm^2 C^{-2}$ is the Coulomb constant

$q_p = 1.6\cdot 10^{-19} C$ is the magnitude of the charge of the proton

$q_e = 1.6\cdot 10^{-19}C$ is the magnitude of the charge of the electron

$r=8\cdot 10^{-11}m$ is the distance between the proton and the electrons

Substituting the values into the formula, we find

$F=(9\cdot 10^9 ) \frac{(1.6\cdot 10^{-19})^2}{(8\cdot 10^{-11})^2}=3.6\cdot 10^{-8} N$

8 0

3 years ago

___is when a force is exerted on an object and the object moves a distance in the direction of the force.

defon

The correct option is A) work.

Work is said to be done by a force when a force is exerted on an object and the the object on which the force is exerted moves in the direction of the force.

The amount of work done is given by the product of the force and the displacement made by the object.

Option (B) is incorrect, since Watt is the unit of power.

Option (C) is wrong, since joule is the unit of work or energy.

Option (D) is wrong, since power is the rate at which work is done.

5 0

3 years ago

Read 2 more answers

A football punter wants to kick the ball so that it is in the air for 4.2 s and lands 55 m from where it was kicked. Assume that

earnstyle [38]

Answer:

$57.24^{\circ}$

24.21 m/s

Explanation:

x = Displacement in x direction = 55 m

y = Displacement in x direction = 0

$y_0$ = Height of the ball when the ball is kicked = 1 m

t = Time taken = 4.2 s

$a_y=g$ = Acceleration due to gravity = $-9.81\ \text{m/s}^2$

u = Initial velocity of ball

Displacement in x direction is given by

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow 55=4.2u_x+0\\\Rightarrow u_x=\dfrac{55}{4.2}\\\Rightarrow u\cos\theta=13.1\ \text{m/s}\\\Rightarrow u=\dfrac{13.1}{\cos\theta}$

Displacement in y direction is given by

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 0=1+4.2u\sin\theta+\dfrac{1}{2}\times (-9.81)\times 4.2^2\\\Rightarrow 0=4.2u\sin\theta-85.5242\\\Rightarrow u\sin\theta=20.36\ \text{m/s}\\\Rightarrow u=\dfrac{20.36}{\sin\theta}$

$\dfrac{13.1}{\cos\theta}=\dfrac{20.36}{\sin\theta}\\\Rightarrow \dfrac{20.36}{13.1}=\dfrac{\sin\theta}{\cos\theta}\\\Rightarrow \theta=\tan^{-1}\dfrac{20.36}{13.1}\\\Rightarrow \theta=57.24^{\circ}$

The ball should be kicked at an angle of $57.24^{\circ}$

$u=\dfrac{13.1}{\cos\theta}=\dfrac{13.1}{\cos57.24^{\circ}}\\\Rightarrow u=24.21\ \text{m/s}$

The initial speed of the ball is 24.21 m/s.

3 0

3 years ago

A straight wire of length 0.56 m carries a conventional current of 0.4 amperes. What is the magnitude of the magnetic field made

djverab [1.8K]

Answer:

The magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.

Explanation:

Given;

length of the straight wire, L = 0.56 m

conventional current, I = 0.4 A

distance of magnetic field from the wire, r = 2.6 cm = 0.026 m

To determine magnitude of magnetic field made by current in the wire, we will apply Bio-Savart Law;

$B = \frac{\mu_o}{4\pi r} \frac{LI}{\sqrt{r^2 +(L/2)^2} } \\\\B = \frac{4\pi *10^{-7}}{4\pi *0.026} \frac{0.56*0.4}{\sqrt{(0.026)^2 +(0.56/2)^2} }\\\\B = 3.846*10^{-6}(0.7966)\\\\B = 3.064*10^{-6} \ T$

Therefore, the magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.

3 0

3 years ago