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3 years ago

Golf Skills pls help mee

Physics

1 answer:

loris [4]3 years ago

6 0

Answer:

1.) Putting club or the putter

2.) Either the 4-, 5-, or the 6- iron club

3.) 14 clubs

4.) The height of the golfer

Explanation:

I used to golf

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A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a

Arlecino [84]

Answer:

Mathematically:

That will be

= 1500 x 15 x 4500

= 101,250,000

5 0

3 years ago

Physics. Need help. Brainlieast answer for most/ all of the answers answered

Mumz [18]

ALL of the following work assumes NO AIR RESISTANCE:

1). an object moving under the influence of only gravity, and not in orbit; its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²

2). a parabola

3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.

4). a). the one that was thrown horizontally; b). both hit the ground at the same time; c). both hit the ground with the same vertical velocity

5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion

6). a). 1.8 seconds; b). 13.1 meters; c). 17.6 m/s down; d). 7.3 m/s; gravity has no effect on horizontal motion

7). 45 m/s

8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand

9). a). 4.49 m/s; b). 29.7 m/s

10). 7.24 meters

11). 700 meters

12). A). 103.7 meters ( ! she's in big trouble ! ); B). 17.5 meters

3 0

3 years ago

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro

viktelen [127]

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$ = 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$ =Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

4 0

3 years ago

Read 2 more answers

A force of 960 newtons stretches a spring 4 meters. A mass of 60 kilograms is attached to the end of the spring and is initially

Drupady [299]

Answer:

x(t) = - 6 cos 2t

Explanation:

Force of spring = - kx

k= spring constant

x= distance traveled by compressing

But force = mass × acceleration

==> Force = m × d²x/dt²

===> md²x/dt² = -kx

==> md²x/dt² + kx=0 ------------------------(1)

Now Again, by Hook's law

Force = -kx

==> 960=-k × 400

==> -k =960 /4 =240 N/m

ignoring -ve sign k= 240 N/m

Put given data in eq (1)

We get

60d²x/dt² + 240x=0

==> d²x/dt² + 4x=0

General solution for this differential eq is;

x(t) = A cos 2t + B sin 2t ------------------------(2)

Now initially

position of mass spring

at time = 0 sec

x (0) = 0 m

initial velocity v= = dx/dt= 6m/s

from (2) we have;

dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)

put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

==> 2B = -6

B= -3

Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get

x(t) = - 6 cos 2t

==>

4 0

3 years ago

HELP NEEDED ASAP I'M GIVING BRAINLIEST

guajiro [1.7K]

Answer:

d . A stream of particles

2. D. Radiowave

3. Microwaves

7 0

3 years ago

Read 2 more answers