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2 years ago

A velocity-time graph is shown below: (2 points)

Physics

1 answer:

BabaBlast [244]2 years ago

5 0

Answer:

2m/s²

Explanation:

V=Vstart+at
rewrite that to find a so a=(V-Vstart)/t

part A of graph

a=(20m/s–0m/s)/5s
a=4m/s²

part B of graph

a=(0m/s–0m/s)/5s
a=0m/s²

the average between the two is both answers added divided by the number of answers

(4m/s²–0m/s²)/2
4m/s²/2
2m/s²

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The diagram below shows a golf ball being struck by a club. The ball leaves the club with a speed of 40 meters per second at an

hjlf

Answer:

61.22m

Explanation:

Maximum height in projectile is expressed as;

H = u²sin²θ/2g

u is the speed

g is the acceleration due to gravity

θ is the angle of projection

Substitute the given values into the formula;

H = u²sin²θ/2g

H = 40²sin²60/2g

H = 1600(0.8660)²/2(9.8)

H = 1,199.9/19.6

H = 61.22m

Hence the maximum height achieved by the ball is 61.22m

4 0

2 years ago

Danny Diver weighs 500 N and steps off a diving board 10 m above the water. Danny hits the water with kinetic energy of

Morgarella [4.7K]

Answer:

Danny hits the water with kinetic energy of 5000 J.

Explanation:

Given that,

The Weight of Danny Diver,

F = 500 N

m*g= 500 N

He steps off a diving board 10 m above the water.

h=10 m

when Danny diver hits water he generates the kinetic energy.

We need to find the kinetic energy of the water.

Let kinetic energy is K.

K = m*g*h

Where g is acceleration due to gravity.

that g= 9.8 m/s^2

now substituting the values in above equation

K= (500) * 10

K= 5000 J

Hence,

he hits the water with kinetic energy of 5000 J.

Learn more about Kinetic energy here:

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7 0

1 year ago

A charge q1= 3nC and a charge q2 = 4nC are located 2m apart. Where on the line passing through these charges is the total electr

Ipatiy [6.2K]

Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = $\frac{9\times 10^9\times3\times10^{-9}}{d^2}$

Field due to charge = 4 X 10⁻⁹ C

$E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$

These two fields will be equal and opposite to make net field zero

$\frac{9\times 10^9\times3\times10^{-9}}{d^2}$ = $[tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$ [/tex]

$\frac{3}{d^2} =\frac{4}{(2-d)^2}$

$\frac{2-d}{d} =\frac{2}{1.732}$

d = 0.928

5 0

3 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

8 0

3 years ago

At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There

katen-ka-za [31]

a) Constant

b) Constant

Explanation:

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

$\tau_{net} = I \alpha$ (1)

where

$\tau_{net}$ is the net torque acting on the object in rotation

I is the moment of inertia of the object

$\alpha$ is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

$\alpha = \frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time interval

So we can rewrite eq.(1) as

$\tau_{net}=I\frac{\Delta \omega}{\Delta t}$

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

$\tau_{net}=0$

From the previous equation, this implies that

$\Delta \omega =0$

Which means that the angular velocity at that instant does not change anymore.

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

$\tau_{net}=0$

Therefore, this means that

$\Delta \omega =0$

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0

3 years ago