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3 years ago

A zebra starts from rest and accelerates at 1.9 m/s2 what will be the acceleration after 5 seconds

Physics

1 answer:

Alex73 [517]3 years ago

3 0

Answer:

d = 23.75 m

The zebra has gone 23.75m after 5 seconds.

Corrected question;

A zebra starts from rest and accelerates at 1.9 m/s2. How far has the zebra gone after 5 seconds.

Explanation:

From the equation of motion;

d = vt + 0.5at^2 .........1

Where;

d = displacement

v = initial velocity = 0

t = time taken= 5 seconds

a = acceleration = 1.9 m/s^2

Substituting the given values into equation 1;

d = 0(5) + 0.5(1.9×5^2)

d = 0.5(1.9×5^2)

d = 23.75 m

The zebra has gone 23.75m after 5 seconds.

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Erica throws a tennis ball against a wall, and it bounces back. Which force is responsible for sending the ball back to Erica?

Anna71 [15]

OPTIONS :

A.) the force that the ball exerts on the wall

B.) the frictional force between the wall and the ball

C.) the acceleration of the ball as it approaches the wall

D.) the normal force that the wall exerts on the ball

Answer: D.) the normal force that the wall exerts on the ball

Explanation: The normal force acting on an object can be explained as a force experienced by an object when it comes in contact with a flat surface. The normal force acts perpendicular to the surface of contact.

In the scenario described above, Erica's tennis ball experiences an opposite reaction after hitting the wall.This is in relation to Newton's 3rd law of motion, which states that, For every action, there is an equal and opposite reaction.

The reaction force in this case is the normal force exerted on the ball by the wall perpendicular to the surface of contact.

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3 years ago

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown

nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

$v_x = 15.0 m/s$

and with an initial vertical velocity of

$v_{y0} = -20.0 m/s$

where we have put a negative sign to indicate that the direction is downward.

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

$y(t) = h + v_{0y} t + \frac{1}{2}gt^2$ (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

$0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m$

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

$v_y = -20 m/s$

So the vertical position of the balloon after a time t is

$y(t) = h + v_y t$

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

$y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m$

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

$v_x = 15.0 m/s$

So the horizontal distance travelled in t = 6.00 s is

$d_x = v_x t = (15.0 m/s)(6.00 s)=90 m$

Considering also that the vertical height of the balloon after t=6.00 s is

$d_y = 176.6 m$

The distance between the balloon and the rock can be found by using Pythagorean theorem:

$d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m$

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer at rest in the basket is

$v_y (t) = gt$

Substituting time t=6.00 s, we find

$v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s$

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer on the ground is now given by

$v_y (t) = v_{0y} + gt$

Substituting time t=6.00 s, we find

$v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s$

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3 years ago

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3 years ago

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3 years ago

A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol

PIT_PIT [208]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a) at 4.75 cm

b) at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula $E = \dfrac{kq }{d}$

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}$

E = 1461.95 N/C

c) The electric field E is calculated as:

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}$

E = 239.76 N/C

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3 years ago