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4 years ago

11

5-A baseball is hit with a velocity of 15 m/s at an angle of 40° in the horizontal direction. What are the horizontal and vertic

al vector components of the velocity vector?
A•Horizontal =9.64 m/s
Vertical=11.49 m/s

B•Horizontal=11.49 m/s
Vertical=9.64 m/s

C•Horizontal=25 m/s
Vertical=25 m/s

D•Horizontal =75 m/s
Vertical =75 m/s

Number 6 is in the picture. TIA

1 answer:

borishaifa [10]4 years ago

5 0

<u>Answer</u>

B•Horizontal=11.49 m/s

Vertical=9.64 m/s

Using the concept of a trigonometric ratios,

sin θ = y/hypotenuse

where y is the vertical component.

sin 40 = y/15

y = 15 × sin 40

= 9.64 m/s

vertical component = 9.64 m/s

cos θ = x/hypotenuse

where x is the horizontal component

cos 40 = x/15

x = 15 × cos 15

=11.49

Horizontal component = 11.49 m/s

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Two protons (each with q = 1.60 x 10-19)

otez555 [7]

Answer:

230.4 N

Explanation:

From the question given above, the following data were obtained:

Charge (q) of each protons = 1.6×10¯¹⁹ C

Distance apart (r) = 1×10¯¹⁵ m

Force (F) =?

NOTE: Electric constant (K) = 9×10⁹ Nm²/C²

The force exerted can be obtained as follow:

F = Kq₁q₂ / r²

F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²

F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰

F = 2.304×10¯²⁸ / 1×10¯³⁰

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3 years ago

As a horse finishes its trot around the corral, it slows from 4m/s to a stop in 3

german

As a horse finishes its trot around the corral, it slows from 4m/s to a stop in 3
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3 years ago

A rock is tossed straight up from the ground with a speed of 21 m/s . When it returns, it falls into a hole 10 m deep.a.) What i

Arte-miy333 [17]

(a) 25.2 m/s

Let's take the initial vertical position of the rock as "zero" (reference height).

According to the law of conservation of energy, the speed of the rock as it reaches again the position "zero" after being thrown upwards is equal to the initial speed of the rock, 21 m/s (in fact, if there is no air resistance, no energy can be lost during the motion; and since the kinetic energy depends only on the speed of the rock:

$K=\frac{1}{2}mv^2$

and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)

This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using

$v^2 = u^2 + 2gd$

where

u = 21 m/s is the initial speed of the rock as it enters the hole

g = 9.8 m/s^2 is the acceleration due to gravity

d = 10 m is the depth of the hole

Substituting,

$v=\sqrt{u^2 +2gd}=\sqrt{(21 m/s)^2+2(9.8 m/s^2)(10 m)}=25.2 m/s$

(b) 4.72 s

The vertical position of the rock at time t is given by

$y(t) = v_y t - \frac{1}{2}gt^2$

where

$v_y = 21 m/s$ is the initial vertical velocity

Substituting y(t)=-10 m, we can then solve the equation for t to find the time at which the rock reaches the bottom of the hole:

$-10 = 21 t - \frac{1}{2}(9.8)t^2\\10+21 t -4.9t^2 = 0$

which has two solutions:

t = -0.43 s --> negative, so we discard it

t = 4.72 s --> this is our solution

7 0

4 years ago

determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund

wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is $a_G= 6.78 m/s^2$

<u>Explanation</u>:

$N_A+N_B-Mg = 0$

$-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0$

$0.8N_B +0.8N_A = 975a_G$

$N_A+N_B = 9564.75$ -------------(1)

$-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0$

$-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0$

$-2.26N_A -0.06N_B= 0$ ----------------(2)

Solving the equation (1) and(2)

$N_A + N_B = 9564.75$

$-2.26N_A-0.06N_B=0$

$N_A = -260.85N$

$N_B = 9825.60N$

$\mu_s N_B + \mu_s N_A = 975a_G$

$0.8(9825.60)+0.8(-260.85) = 975a_G$ $a_G=\frac{7651.8}{975}$ $a_G_1=7.4848m/s^2$

Next lets assume that the front wheels contact with the ground N_A = 0

$F_B = Ma_G$

$N_B = M_g$

$N_B - M_g = 0$

$N_B(b-a) –F_Bh = 0$

$F_B = 975a_G$

$N_B-975(9.8) = 0$

$N_B=9564.75N$

$9564.75(2.20 -1.82) -F_B(0.55)=0$

$\frac{3634.605}{0.55}=F_B$

$F_B = 6608.3$

$F_B = Ma_G$

$6608.3 = 975a_G$

$a_G = 6.7778 m/s^2$

$a_G_2 = 6.78m/s^2$

Choosing the critical case

$a_G = min(a_G_1 ,a_G_2)$

$a_G = min(7.848, 6.78)$

$a_G= 6.78 m/s^2$

3 0

4 years ago

A vertical rectangular wall with a width of 20 m and a height of 12 m is holding a 7-m-deep water body. The resultant hydrostati

denis23 [38]

Answer:

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Explanation:

Force can be expressed as;

F = ma

F = mg ......1

Where,

F = force

m = mass

a = acceleration

g = acceleration due to gravity

And we know that mass of an object is the product of

Density and Volume

m = pV

And volume V = area × height = Ah

m = pAh .....2

Substituting m into equation 1

F = pAhg = pghA ......3

Given;

Density of water p = 1000kg/m^3

g = 9.81m/s^2

Area A = 20m × 12m = 240m^2

h = depth/2 = 7/2 = 3.5m

Substituting the values into equation 3, we have;

F = 1000×9.81×3.5×240

F = 8240400 N

Resultant hydrostatic force F = 8240 kN

5 0

3 years ago