A top-fuel dragster accelerates from rest to a velocity of 100 m/s in 8 s. What is the acceleration?

Jack and Jill have made up since the previous HW assignment, and are now playing on a 10 meter seesaw. Jill is sitting on one en

Airida [17]

Answer: 3 m.

Explanation:

Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by gravity acting on both children must be 0.

As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.

If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):

mJill* 5m -mJack* d = 0

60 kg*5 m -100 kg* d =0

Solving for d:

d = 3 m.

6 0

2 years ago

A 1500 kg truck is acted upon by a force that decreases its speed from 25 m/s to 15 m/s in 8 s. What is the magnitude of the for

Citrus2011 [14]

Answer:

F = 1,875 N

Explanation:

force=

$\frac{change \: in \: momentum}{time \: taken}$

∆H = m∆V

where ∆H ----> change in momentum.

( final momentum - initial momentum )

and ∆V ----> change in velocity

( final velocity - initial velocity )

and m ----> is mass

then f =

$\frac{1500 \times (25 - 15)}{8}$

= 1,875 N

4 0

2 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance during a time t, and object B falls th

UNO [17]

Answer:

Distance covered by B is 4 times distance covered by A

Explanation:

For an object in free fall starting from rest, the distance covered by the object in a time t is

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

where

s is the distance covered

g is the acceleration due to gravity

t is the time elapsed

In this problem:

- Object A falls through a distance $s_A$ during a time t, so the distance covered by object A is

$s_A=\frac{1}{2}gt^2$

- Object B falls through a distance $s_B$ during a time 2t, so the distance covered by object B is

$s_B=\frac{1}{2}g(2t)^2 = 4(\frac{1}{2}gt^2)=4s_A$

So, the distance covered by object B is 4 times the distance covered by object A.

5 0

2 years ago

How many electrons must be remowel from an electricaly Nurutral Silvor Coin to five it a charge of 3.2 NC ?

kakasveta [241]

Answer:

#_electrons = 2 10¹⁰ electrons

Explanation:

For this exercise we can use a direct rule of three proportions rule. If an electron has a charge of 1.6 10⁻¹⁹ C how many electrons have a charge of 3.2 10⁻⁹ C

#_electrons = 3.2 10⁻⁹ ( $\frac{1}{1.6 \ 10^{-19}}$ )

#_electrons = 2 10¹⁰ electrons

7 0

2 years ago

A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l

mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be $\who_w = 1000 kg/m^3$

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

$F_s + F_b = W$

Where $F_s = kx$ N is the spring force, $F_b = W_w = m_wg = \rho_w V_s g$ is the buoyancy force, which equals to the weight $W_w$ of the water displaced by the submerged portion of the cylinder, which is the product of water density $\rho_w$ , submerged volume $V_s$ and gravitational constant g. W = mg is the weight of the metal cylinder.