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LuckyWell [14K]
4 years ago
16

The function h(t) = –16t2 + 96t + 6 represents an object projected into the air from a cannon. The maximum height reached by the

object is 150 feet. After how many seconds does the object reach its maximum height? 2 seconds 3 seconds 6 seconds 9 seconds
Physics
2 answers:
Rainbow [258]4 years ago
6 0

Answer:

3 seconds

Explanation:

Since h(t) represents the height and t represents the time, we can set the equation equal to 150 to find t.

-16t^2+96t+6=150

Subtract 150 from both sides to set the equation equal to 0, to find the solutions.

-16t^2+96t-144=0

Factor out -16, because all of the terms are divisible by it.

-16(t^2+6t+9)=0

Now we can focus on the terms inside the parenthesis and factor it again.

t^2-6t+9=0

We need to find two value that can be multiplied to get 9 and added to get -6.

-3 and -3 works.

Thus, we get (x-3)(x-3).

Now solve for 0.

x-3=0

x=3

The object reaches its maximum height after 3 seconds.

svp [43]4 years ago
6 0

Answer:

3 seconds

Explanation:

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A motorbike travels 100m in 2.5 seconds. What is its speed?
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3 years ago
In a meeting of mimes, mime 1 goes through a displacement d1 = (6.00 m)i + (5.74 m)j and and mime 2 goes through a displacement
Blizzard [7]

a) Vector product: |d1 × d2| = (34.2 m) k

b) Scalar product: d1 · d2 = -0.874 m

c) (d1 + d2) · d2 = 16.1 m

d) Component of d1 along direction of d2: -0.21 m

Explanation:

a)

In this part, we want to calculate

|d1 × d2|

Which is the vector product between the two displacements d1 and d2.

The two vectors are:

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

The vector product of two vectors (a_x,a_y,a_z) and (b_x,b_y,b_z) is also a vector which has components:

r=(r_x,r_y,r_z)\\r_x=a_yb_z-a_zb_y\\r_y=a_zb_x-a_xb_z\\r_z=a_xb_y-a_yb_x

We notice immediatly that in this problem ,the two vectors d1 and d2 lie in the x-y plane, so they do not have components in zero. Therefore, the vector product has only one component, which is the one in z, and it is:

r_z=(6.00)(2.9)-(5.74)(-2.92))=34.2 m

Therefore, the vector product of d1 and d2 is:

|d1 × d2| = (34.2 m) k

b)

In this case, we want to calculate

d1 · d2

Which is the scalar product between the two displacements.

The scalar product of two vectors (a_x,a_y,a_z) and (b_x,b_y,b_z) is a scalar given by:

a \cdot b = a_x b_x + a_y b_y + a_z b_z

In this problem,

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

Therefore, the scalar product between the two vectors is:

d_1 \cdot d_2 = (6.00)(-2.92)+(5.74)(2.9)=-0.87m

c)

In this  case, we want to calculate

(d1 + d2) · d2

Which means that first we have to calculate the resultant displacement d1 + d2, and then calculate the scalar product of the resultant vector with d2.

Given two vectors  (a_x,a_y,a_z) and (b_x,b_y,b_z), the resultant vector is also a vector given by

r=(r_x,r_y,r_z)\\r_x=a_x+b_x\\r_y=a_y+b_y\\r_z=a_z+b_z

In this case,

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

So the resultant vector is

r_x=6.00+(-2.92)=3.08 m\\r_y = 5.74+2.9=8.64 m

So

(d_1+d_2)=(3.08 m,8.64 m)

And calculating the scalar product with d2, we find:

(d_1 + d_2)\cdot d_2 = (3.08)(-2.92)+(8.64)(2.9)=16.1 m

d)

The component of a vector a along another vector b is given by

a_b = \frac{a\cdot b}{|b|}

wherea\cdot b is the scalar product between and b

|b| is the magnitude of vector b

In this problem, we have the two vectors

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

We want to find the component of d1 along the direction of d2.

We already calculated the scalar product of the two vectors in part b):

d1 · d2 = -0.874 m

The magnitude of a vector b is given by

|b|=\sqrt{b_x^2+b_y^2+b_z^2}

So, for vector d2,

|d_2|=\sqrt{(-2.92)^2+(2.9)^2}=4.1 m

Now we can calculate the component of d1 along d2:

d_1_{d_2}=\frac{d_1 \cdot d_2}{|d_2|}=\frac{-0.874}{4.1}=-0.21 m

Learn more about operations with vectors:

brainly.com/question/2927458

brainly.com/question/2088577

brainly.com/question/1592430

#LearnwithBrainly

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A pulley really saves effort when you have more than one pulley working together. By looping a rope around two, three, or even four pulleys, you can really cut down on the effort needed to lift something. The trade-off? Well, as you increase the number of pulleys, you also increase the distance you have to pull the rope. In other words, if you use two pulleys, it takes half the effort to lift something, but you have to pull the rope twice as far. Three pulleys will result in one-third the effort — but the distance you have to pull the rope is tripled!

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