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3 years ago

14

Evan drives a golf ball with an initial velocity of 49 meters per second at an angle of 16 degrees own a flat driving range. How

far will the golf ball land?

1 answer:

alexandr1967 [171]3 years ago

5 0

Answer:

<h2>solusion </h2><h2>Explanation:</h2><h2>the answer in 984 </h2>

tha

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A wire of cross-sectional area 5.00 106 m2 has a resistance of 1.75 O. What is the resistance of a wire of the same material and

MakcuM [25]

Answer:

the resistance of the second wire is 1 ohm.

Explanation:

Given;

cross sectional area of the first wire, A₁ = 5.00 x 10⁶ m²

resistance of the first wire, R₁ = 1.75 ohms

cross sectional area of the second wire, A₂ = 8.75 x 10⁶ m²

resistance of the second wire, R₂ = ?

The resistance of a wire is given as;

R ∝ $\frac{L}{A}$

Since the length of the two wires is constant

R₁A₁ = R₂A₂

$R_2 = \frac{R_1A_1}{A_2} \\\\R_2 = \frac{1.75\ \times \ 5.00\times 10^6}{8.75\times 10^6} \\\\R_2 = 1 \ ohm$

Therefore, the resistance of the second wire is 1 ohm.

6 0

3 years ago

A 0.030 kg lead bullet hits a steel plate, both initially at 20?C. The bullet melts and splatters on impact. Assume that 80% of

Dmitry_Shevchenko [17]

Answer:

(a). The required is 1871.2 J.

(b). The speed the lead bullet is 395 m/s.

(c). The 20% of energy must have gone into collision between steel plate and bullet.

Explanation:

Given that,

Mass of bullet = 0.030 kg

Temperature = 20°C

(a). We need to heat required to increase the temperature of the lead bullet and melt it

Using formula of heat

$Q=mS\Delta T+mL$

Where, m = mass of lead bullet

S = specific heat

L = latent heat

T = Temperature

Put the value into the formula

$Q=0.030\times130\times(327.5-20)+0.030\times22400$

$Q=1871.2\ J$

The required is 1871.2 J.

(b). Assume that 80% of the bullet’s kinetic energy goes into increasing its temperature and then melting it

We need to calculate the energy

$Q=\dfrac{1871.2}{0.8}$

$Q=2339 J$

We need to calculate the speed the lead bullet

Using formula of speed

$K.E=Q$

$\dfrac{1}{2}mv^2=2339$

$v^2=\dfrac{2339\times2}{0.030}$

$v=\sqrt{\dfrac{2339\times2}{0.030}}$

$v=394.8 = 395\ m/s$

The speed the lead bullet is 395 m/s.

(c). The 20% of energy must have gone into collision between steel plate and bullet.

Hence, This is the required solution.

6 0

3 years ago

A man tries to push a 200 kg Car that moves at a acceleration 0.50 m/s2. The man is able to displace the car 10 m. How much work

yawa3891 [41]

The work done by the man pushing the car over the given distance is 1000J.

Given the data in the question;

Mass of car; $m = 200kg$

Acceleration of the car; $a = 0.5m/s^2$

Distance covered by the car; $d = 10m$

Work done; $W = \ ?$

<h3>Work done</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

$Work\ done = f * d$

Where f is force applied and d is distance travelled.

To determine the work done by the man, we first solve for the force applied F.

From Newton's Second Law; $Force \ F = m * a$

We substitute our given values into the expression

$F = m * a \\\\F = 200kg * 0.5m/s^2\\\\F = 100kg.m/s^2$

Next we substitute our values into the expression of work done above.

$Work \ done = f * d\\\\Work \ done = 100kg.m/s^2 * 10m\\\\Work \ done = 1000kgm^2/s^2\\\\Work \ done = 1000J$

Therefore, the work done by the man pushing the car over the given distance is 1000J.

Learn more about work done: brainly.com/question/26115962

7 0

2 years ago

When a honeybee flies through the air, it develops a charge of +20 pC . Part A How many electrons did it lose in the process of

Yuri [45]

Answer:

1.3 × 10⁸ e⁻

Explanation:

When a honeybee flies through the air, it develops a charge of +20 pC = + 20 × 10⁻¹² C. This is a consequence of losing electrons (negative charges). The charge of 1 mole of electrons is 96468 C (Faraday's constant). The moles of electrons representing 20 pC are:

20 × 10⁻¹² C × (1 mol e⁻/ 96468 C) = 2.1 × 10⁻¹⁶ mol e⁻

1 mole of electrons has 6.02 × 10²³ electrons (Avogadro's number). The electrons is 2.1 × 10⁻¹⁶ moles of electrons are:

2.1 × 10⁻¹⁶ mol e⁻ × (6.02 × 10²³ e⁻/ 1 mol e⁻) = 1.3 × 10⁸ e⁻

7 0

4 years ago

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. Yo

BigorU [14]

Answer:

$\lambda = 6.25\times10^{-9}$ = 625 nm

Explanation:

We now that for

for maximum intensity(bright fringe) d sinθ=nλ n=0,1,2,....

d= distance between the slits, λ= wavelength of incident ray

for small θ, sinθ≈tanθ= y/D where y is the distance on screen and D is the distance b/w screen and slits.

Given

d=1.19 mm, y=4.97 cm, and, n=10, D=9.47 m

applying formula

λ= (d*y)/(D*n)

putting values we get

$\lambda = \frac{1.19\times10^{-3}\times4.97\times10^{-2}}{9.47\times10}$

on solving we get

$\lambda = 6.25\times10^{-9}$ = 625 nm

8 0

3 years ago