Mr. Dunn drives 64.8km from work at a speed of 48km/h. Mrs. Dunn drives 81.2km from work

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

Ostrovityanka [3204]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

This gives us our initial equation.

Next, we need to examine the vertical trajectory.

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Utilizing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Now let’s solve for t.

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

The computed velocity magnitude is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The ball's angle is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

4 0

3 months ago

Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a

Yuliya22 [3333]

Answer:

a)106.48 x 10⁵ kg.m²

b)144.97 x 10⁵ kgm² s⁻¹

Explanation:

a)Given

m = 5500 kg

l = 44 m

The moment of inertia for one blade

$I$ = 1/3 x m l²

where m denotes the mass of the blade

l represents the length of each blade.

Substituting the necessary values, the moment of inertia for one blade is

$I$ = 1/3 x 5500 x 44²

$I$ = 35.49 x 10⁵ kg.m²

Total moment of inertia for 3 blades

$I$ = 3 x 35.49 x 10⁵ kg.m²

$I$ = 106.48 x 10⁵ kg.m²

b) The angular momentum 'L' is calculated using

L = $I$ x ω

where,

$I$ = the moment of inertia of the turbine i.e 106.48 x 10⁵ kg.m²

ω= angular velocity =2π f

f represents the frequency of rotation of the blade i.e 13 rpm

f = 13 rpm=>= 13 / 60 revolutions per second

ω = 2π f => 2π x 13 / 60 rad / s

L= $I$ x ω =>106.48 x 10⁵ x 2π x 13 / 60

= 144.97 x 10⁵ kgm² s⁻¹

7 0

1 month ago

A car drives 215 km east and then 45 km north. What is the magnitude of the car’s displacement? Round your answer to the nearest

Ostrovityanka [3204]

The Pythagorean Theorem can be utilized here: Imagine a car navigating through traffic—when it turns left to travel north, a right angle of 90 degrees is formed. However, the displacement is always the shortest distance connecting the origin and the endpoint, which forms a triangle in this scenario. In a right triangle, the Pythagorean theorem applies: 215^2+45^2=c^2; therefore, v=√(215^2+45^2).

8 0

1 month ago

Read 2 more answers

A nerve signal is transmitted through a neuron when an excess of Na+ ions suddenly enters the axon, a long cylindrical part of t

Yuliya22 [3333]

Answer:

1.32.225 N/C, moving away from the point charge

2. 8.972*10^-12 C

3. the field is oriented away from the axon

Explanation:

The calculation for the electric field is illustrated below:

E = k*|q|/r²

Where:

E = electric field; k = 8.98755*10⁹ N*m²/C²; r = distance separating the field being measured from the point charge = 0.05 m; q = point charge

For a length of 0.100 m of the axon, the value of q is calculated as:

q = (5.6*10¹¹)*(+e)*(0.001)

+e = charge of an electron = 1.60217*10^-19 C

Therefore:

q = (5.6*10¹¹)*(1.60217*10^-19)*(0.0001) = 8.972*10^-12 C

Consequently:

E = (8.98755*10⁹)*(8.972*10^-12)/0.05² = 32.255 N/C

A positive point charge produces an electric field that radiates outward, while a negative point charge creates an electric field directed inward.

3 0

2 months ago

A rocket takes off vertically from the launch pad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At

kicyunya [3294]

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + $\frac{1}{2}$ a $t^{2}$

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + $\frac{1}{2}$ (2.25 x $15.4^{2}$ )

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s² (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

$v^{2}$ = $u^{2}$ + 2gs

$0^{2}$ = $34.65^{2}$ + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m + 61.19 m = 328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying $v^{2}$ = $u^{2}$ + 2gs

$v^{2}$ = $0^{2}$ + 2(9.81 x 328)

$v^{2}$ = 0 + 6435.36

v = $\sqrt{6435.36}$ = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

7 0

2 months ago