1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ziro4ka [17]
3 years ago
8

h(t) = - 16t2 + 64t + 112 where t is the time in seconds. After how many seconds does the arrow reach it maximum height? Round t

o the nearest tenth of a second if necessary. The arrow reaches its maximum height in seconds.
Physics
1 answer:
laila [671]3 years ago
8 0

Answer:

2 seconds

Explanation:

The function of height is given in form of time. For maximum height, we need to use the concept of maxima and minima of differentiation.

h(t)=-16t^{2}+64t+112

Differentiate with respect to t on both the sides, we get

\frac{dh}{dt}=-32t+64

For maxima and minima, put the value of dh / dt is equal to zero. we get

- 32 t + 64 = 0

t = 2 second

Thus, the arrow reaches at maximum height after 2 seconds.

You might be interested in
Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
arsen [322]

Answer:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates  to find the horizontal distance traveled by the electron when it hits the plate.

acceleration a=qE/m=1.6*10^{-19}*4*10^3/9.1*10^{-31} =0.7*10^{15}=7*10^{14} m/s

now we find the horizontal distance traveled by electrons hit the plates

horizontal distance

X=u[2y/a]^{1/2}

=4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}

=3*10^{-2}= 3 cm

5 0
4 years ago
A missile is launched vertically from a missile silo, it will explode after 32 s. It's launch speed was 145 m/s, and it doesn't
Karolina [17]

Answer:

Landed before it explodes

Explanation:

vf = vi + at,

0 = 145 - (9.8)t,

t = 14.79 s (Time to reach highest point)

14.79 x 2 = 29.59 s (Time to land on the ground)

It will have landed before it explodes because both the time to reach the highest point and the time to land on the ground are less than 32 seconds.

4 0
3 years ago
A block of mass 500g is pulled from rest on ahorizontal frictionless bench by a steady force F and travels 8m in 2s find the acc
Korvikt [17]

<u>We are Given:</u>

Mass of the block (m) = 500 grams or 0.5 Kg

Initial velocity of the block (u) = 0 m/s

Distance travelled by the block (s) = 8 m

Time taken to cover 8 m (t)= 2 seconds

Acceleration of the block (a) = a m/s²

<u>Solving for the acceleration:</u>

From the seconds equation of motion:

s = ut + 1/2* (at²)

<em>replacing the variables</em>

8 = (0)(2) + 1/2(a)(2)²

8 = 2a

a = 4 m/s²

Therefore, the acceleration of the block is 4 m/s²

3 0
3 years ago
You slide a hockey puck across the ice in a hockey rink
REY [17]

Answer-

mk thats kool.

8 0
3 years ago
Read 2 more answers
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
4 years ago
Other questions:
  • If a plane can travel 450 miles per hour with the wind and 410 miles per hour against the wind, find the speed of the plane with
    13·2 answers
  • An AC voltage is applied to a purely capacitive circuit. Just as the applied voltage is crossing the zero axis going negative, w
    5·1 answer
  • Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilo
    8·1 answer
  • Which of the following statements concerning the lab is TRUE?
    7·1 answer
  • The bleeder is opened at a caliper and no brake fluid flows out. Technician A says that this is normal and that brake fluid shou
    9·1 answer
  • What happens to momentum during a collision?...i give brainliest
    9·1 answer
  • The problem of perception is best characterized as
    15·1 answer
  • 3. Light travels from the Sun to Earth in 8.3 min. Given that the speed of light is 3.00108 m/s, what is the distance in meters
    10·1 answer
  • 2. A train is traveling at 75 km/hour. How far can it travel in 20 hours?
    5·2 answers
  • If a big person and a small person run up the stairs at the same time who has more force
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!