A tow truck pulls a 1,500-kilogram car with a net force of 4,000 newtons. What is the acceleration of the car?

At a meeting of physics teacher in Montana, the teachers were asked to calculate where a flour sack would land if dropped from a

Harlamova29_29 [7]

At a distance of 469.2 m from the original point below the airplane.

Explanation:

First of all, we have to calculate the time it takes for the sack to reach the ground.

To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

s = 300 m is the vertical displacement

u = 0 is the initial vertical velocity

t is the time

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the it takes for the sack to reach the ground:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(300)}{9.8}}=7.82 s$

Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:

$v_x = 60 m/s$

We also know the total time of flight,

t = 7.82 s

Therefore, we can find the horizontal distance travelled by the sack:

$d=v_x t = (60)(7.82)=469.2 m$

So, the sack will land 469.2 m from the original point below the airplane.

Learn more about free fall and projectile motion:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

brainly.com/question/8751410

#LearnwithBrainly

7 0

2 years ago

Express the measurement 0.00000575 into scientific notation.

g100num [7]

Answer: = 5.75 × 10 -6

Explanation:

= 5.75 × 10-6

(scientific notation)

= 5.75e-6

(scientific e notation)

= 5.75 × 10-6

(engineering notation)

(millionth; prefix micro- (u))

= 0.00000575

(real number)

7 0

3 years ago

An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result

mylen [45]

The distance covered by the acorn is 3.136 m.

<u>Explanation:</u>

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

$s=ut+\frac{1}{2}gt^{2}$

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

$s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}$

So the distance covered by the acorn is 3.136 m.

8 0

3 years ago

In a pendulum system, when is it possible for the potential and kinetic energies both to be equal to zero

muminat

That's ONLY true when the pendulum is hanging
in the center position and not moving.

8 0

3 years ago

A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106

LenKa [72]

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

$E_2 = \frac{E_1q_1}{q_2}$

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

$E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C$

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

8 0

3 years ago