A tennis ball moves 16 meters northward, then 22 meters southward, then 12 meters northward, and finally 32 meters southward.

An atomic scientist was studying an atom and through experimentation found that the total negative charge of the area surroundin

Mars2501 [29]

Answer:

c hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

Explanation:

gerhr fdhdrhedrzjhdrtuhe5eru

4 0

3 years ago

A 12.0 V car battery is being used to power the headlights of a car. Each of the two headlights has a power rating of 41.8 Watts

Marina CMI [18]

Answer:

n = 4.35 x 10¹⁹

Explanation:

Given that

Voltage V= 12 V

Power rating of headlights = 41.8 W

We know that headlights of the car is connected in parallel connection.

We know that

Power = Current x Voltage

P = V I

$I=\dfrac{41.8}{12}\ A$

I=3.48 A

Therefore the total current will be

I'= 3.48 + 3.48 A

I = 6.96 A

We know that

Charge = Current x time

q= I' t

q= 6.96 x 1

q= 6.96 C

The charge on electron ,e= 1.6 x 10⁻¹⁹

q= n e

n=Number of electron

$n=\dfrac{q}{e}$

$n=\dfrac{6.96}{1.6\times 10^{-19}}$

n = 4.35 x 10¹⁹

5 0

3 years ago

An amateur golfer swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for o

Rudiy27

Answer:

The average force ≅ 519.44 N.

Explanation:

Impulse = change in momentum of a body

i.e Ft = m(v - u)

where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.

m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s

So that,

F x 0.00360 = 0.055(34 - 0)

F x 0.00360 = 0.055 x 34

= 1.87

F = $\frac{1.87}{0.0036}$

= 519.4444

The average force exerted on the ball by the club is approximately 519.44 N.

4 0

3 years ago

The main difference between heat and temperature is that temperature is solely dependent on the?

WITCHER [35]

Velocity of molecules within a body

3 0

3 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

3 years ago