Ask question

Ask question

All categories

skelet666 [1.2K]

3 years ago

12

The height h of a projectile launched from ground level is a function of the time t it is in the air. The height in feet for t s

econds is given by the function h(t) = −16t2 + 80t. What is the domain of the function?

1 answer:

il63 [147K]3 years ago

7 0

Answer:6

Explanation:

gghh

You might be interested in

Compared to the size and density of Earth, the Moon has a A. smaller diameter and lower density B. smaller diameter and higher d

MrRa [10]

Answer:

(a) Moon has smaller diameter and lower density than earth

Explanation:

Moon has smaller diameter than earth earth has a diameter of 12742 km whereas moon has a diameter of 3474.21 km

So for density density is given by $density=\frac{mass}{volume}$

From the relation we can see that density is dependent on mass and volume directly proportional to mass and inversely proportional to diameter

But in case of earth and moon mass is more dominating than volume so density is less for moon than earth

6 0

3 years ago

a rugby player passes the ball 5.34 m across the field, where it is caught at the same height as it left his hand. at what angle

MakcuM [25]

Answer:

$31.035^{\circ}$

Explanation:

x = Displacement in x direction = 5.34 m

t = Time taken to travel the displacement

y = Displacement in y direction = 0

u = Initial velocity of ball = 7.7 m/s

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Displacement in x direction is given by

$x=u\cos\theta t\\\Rightarrow t=\dfrac{5.34}{7.7 \cos\theta}$

Displacement in y direction is given by

$y=u\sin\theta t-\dfrac{1}{2}gt^2\\\Rightarrow 0=7.7\sin\theta \dfrac{5.34}{7.7\cos\theta}-\dfrac{1}{2}\times 9.81 (\dfrac{5.34}{7.7\cos\theta})^2\\\Rightarrow 0=7.7\sin\theta-4.905\times \dfrac{5.34}{7.7\cos\theta}\\\Rightarrow 0=7.7^2\sin\theta \cos\theta-4.905\times 5.34\\\Rightarrow 0=7.7^2\dfrac{\sin2\theta}{2}-4.905\times 5.34\\\Rightarrow 0=7.7^2\sin2\theta-4.905\times5.34\times 2\\\Rightarrow \sin2\theta=\dfrac{4.905\times 5.34\times 2}{7.7^2}\\\Rightarrow 2\theta=\sin^{-1}\dfrac{4.905\times 5.34\times 2}{7.7^2}$

$\Rightarrow \theta=\dfrac{62.07}{2}\\\Rightarrow \theta=31.035^{\circ}$

The angle at which the ball was thrown is $31.035^{\circ}$ .

7 0

3 years ago

The first moment of the total cross-sectional area taken about the neutral axis must be?

bekas [8.4K]

The first moment of the total cross sectional area taken about the neutral axis must be zero.

As with non-composite beams, the neutral axis (NA) is the location where the bending stress is zero. The location of the NA depends on the relative stiffness and size of each of the material sections.

Generally, the NA location is determined relative to the bottom surface of the beam. However, this is not mandatory, and the location can be relative to any location. If the bottom is used, then the NA axis is a distance "h"

The distance h can be determined by recalling that the stresses through the cross section must be in equilibrium.

learn more about neutral axis from here: brainly.com/question/28167877

#SPJ4

7 0

2 years ago

Read 2 more answers

angela uses a force of 25 newtons to lift her grocery bag while doing 50 joules of work. how far did she lift the grocery bags

Evgen [1.6K]

We know that:
W=Fs
50J=25N*s
s=50J/25N
s=2m

5 0

3 years ago

Objects 1 and 2 attract each other with a electrostatic force of 36.0 units. If the charge of Object 1 is doubled AND the distan

melamori03 [73]

Answer:

8 units

Explanation:

F=(k*q1*q2)/(r^2)

K is a constant, q1 is charge of 1, q2 is charge of 2, r is distance between the two.

6 0

3 years ago