A sack of groceries with a mas of 22 kg is lifted off the floor with a velocity of 6 m/s. What is the kinetic energy of the sack

I measured my pet dog (Mickey) to be able to run the length of my backyard 500 in 12 seconds. At this rate, how many miles can m

Georgia [21]

The dog runs 300240 miles.

Rate refers to the pace at which something happens. We know that the easiest way to obtain the rate is by the use of proportion. Now we have been told that the dog is able to run 500 miles in 12 seconds.

If the dog runs 500 miles in 12 seconds

The dog runs x miles in 1 second

x = 500 * 1/12

x = 41.7 miles

In two hours we have 2 * 60 * 60 = 7200 seconds

If the dog runs 41.7 miles in 1 second

The dog runs x miles in 7200 seconds

x = 41.7 miles * 7200 seconds/ 1 second

x = 300240 miles

Learn more about proportion:brainly.com/question/2548537

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4 0

1 year ago

A projectile is launched at an angle of 60 degrees to the horizontal at an initial speed of 10 meters per second. What is the ma

Flura [38]

Answer:

5 sq. root 3

Explanation:

theta= 60°

=> u sin theta = 10 × sin 60

= 10× sq. root 3/2

= 5 sq. root 3

3 0

2 years ago

Read 2 more answers

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

2 years ago

Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the bat

RSB [31]

Answer:

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

$C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\$

where;

V is the potential difference between the plates

The charge on the plates is given as;

$Q = \frac{V\epsilon _0 A}{d}$

The energy stored in the capacitor is given as;

$E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2$

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

3 0

2 years ago

a ball is projected horizontally with a velocity of 5 m per second from the top of a building 19.6 m high how long will the ball

zepelin [54]

Answer:

1.98s

Explanation:

The time taken to hit the ground is given by

h=ut+ 1/2 at^2

but u =0

so we have

h=1/2at^2

making t the subject

t=√2h/g

√2×19.6/10

1.98s

8 0

2 years ago