Scientists think the is a solid layer of iron with a liquid layer of iron around it

A projectile is launched on earth at an angle θ, relative to horizontal direction. At half of its maximum height the speed of th

Tamiku [17]

Answer:

the angle is about 67.79 degrees

Explanation:

We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)

We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :

$half\,\,max-height = \frac{v_{yi}^2}{4\,g}$

we can use this information to find the y component of the velocity at that height via the formula:

$v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}$

Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:

$1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\$

Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:

$tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o$

3 0

2 years ago

A baseball approaches home plate at a speed of 44.0 m/s, moving horizontally just before being hit by a bat. The batter hits a p

Luda [366]

Explanation:

It is given that,

Speed of the baseball, u = 44 m/s

Speed of the baseball, v = 53 m/s

Mass of the ball, m = 145 g = 0.145 kg

Time of contact between the ball and the bat, t = 2.2 ms = 0.0022 s

$F=ma$

$F=\dfrac{mv}{t}$

$F_1=\dfrac{0.145\ kg\times 44\ m/s}{0.0022\ s}$

F₁ = 2900 N...........(1)

$F=ma$

$F=\dfrac{mv}{t}$

$F_2=\dfrac{0.145\ kg\times 53\ m/s}{0.0022\ s}$

F₂ = 3493.18 N.........(2)

In average vector form force is given by :

$F=F_1+F_2$

$F=(2900i+(-3493.18)\ N$

$F=(2900i-3493.18j)\ N$

Hence, this is the required solution.

6 0

3 years ago

Find the energy released when there is a decrease of 0.3kg of material in a nuclear reaction

MrRissso [65]

Answer:

E = 2.7 x 10¹⁶ J

Explanation:

The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:

$E = mc^2$

where,

E = Energy Released = ?

m = mass of material reduced = 0.3 kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (0.3\ kg)(3\ x\ 10^8\ m/s)^2$

E = 2.7 x 10¹⁶ J

4 0

2 years ago

1.) If you make a false statement or commit a forgery about your motor vehicle insurance you can be guilty of a _____ degree mis

krok68 [10]

A person is guilty of second degree misdemeanor if he or she makes a false statement or commits forgery about their motor vehicle’s insurance.

You would be required to have bodily injury liability insurance in Florida if you are involved in a crash where your vehicle has caused damage to the property of others.

Though, a person is not required to have a Bodily Injury Liability (BI) to legally drive an automobile in the state of Florida. Nevertheless, it would be possible if you are driving under the influence since you are driving while impaired, driving while intoxicated or drunk driving.

6 0

3 years ago

Read 2 more answers

10points asap A force of 30 N acts upon a 7 kg block. Calculate its acceleration.

nekit [7.7K]

Hello! Assuming that the only force acting on the mass is 30N...

Fnet = 30N
Fnet = ma (mass x acceleration)
ma = 30N
a = 30N / m
a = 30N / 7kg
a = 4.2857 m/s^2
a = 4 m/s^2

I hope this helps!

5 0

2 years ago