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3 years ago

What was used as a basket at each end of the gym?

Physics

2 answers:

UNO [17]3 years ago

3 0

Apple basket hope this helped

a_sh-v [17]3 years ago

3 0

Answer: apple basket ✨

Explanation:

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The radius of Venus (from the center to just above the atmosphere) is 6050 km (6050✕103 m), and its mass is 4.9✕1024 kg. An obje

Alecsey [184]

Answer:

(a) The initial speed required is 13116 m/s

(b) The escape speed is 10394 m/s

This problem involves the application of newtons laws of gravitation. The forces in action here are conservative and as a result mechanical energy is conserved.

The full calculation can be found in the attachment below.

Explanation:

In both parts (a) and (b) the energy conservation equation were used. Assumption was made that when the object is very far from the planet the distance from the planet's center approaches infinity and the gravitational potential energy approaches zero.

The calculation can be found below.

6 0

3 years ago

4. A light string is attatched to a heavy rope, and the whole thing is pulled tight. A wave is sent along the light string. When

ale4655 [162]

Answer:

The correct answer to the question is (A)

When it hits the heavy rope, compared to the wave on the string, the wave that propagates along the rope has the same (A) frequency

Explanation:

The speed of a wave in a string is dependent on the square root of the tension ad inversely proportional to the square root of the linear density of the string. Generally, the speed of a wave through a spring is dependent on the elastic and inertia properties of the string

$v = \sqrt{ \frac{T}{\mu } } = \sqrt{ \frac{T}{m/L } }$

Therefore if the linear density of the heavy rope is four times that of light rope the velocity is halved and since

v = f×λ therefore v/2 = f×λ/2

Therefore the wavelength is halved, however the frequency remains the same as continuity requires the frequency of the incident pulse vibration to be transmitted to the denser medium for the wave to continue as the wave is due to vibrating particles from a source for example

7 0

3 years ago

While driving, your car has an initial position of 3.2 m, an initial velocity of -8.4 m/s, and

KIM [24]

Answer:

The position of the car at t = 1.5 s is at -8.1625 meters

Explanation:

The initial position of the car is 3.2 meters

The initial velocity is -8.4 m/s

The constant acceleration is 1.1 m/s²

We need to find the final position of the car at the time t = 1.5 seconds

The displacement s = final position - initial position

$s=ut+\frac{1}{2}at^{2}$ , where u is the initial velocity, a is the

constant acceleration and t is the time

So we can find the final velocity by using the rule:

final position - initial position = $ut+\frac{1}{2}at^{2}$

initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s

Substitute these values in the rule

final position - 3.2 = $(-8.4)(1.5)+\frac{1}{2}(1.1)(1.5)^{2}$

final position - 3.2 = -12.6 + 1.2375

final position - 3.2 = -11.3625

add 3.2 for both sides

final position = -8.1625

That means the car is at 8.1625 meters in opposite direction

The position of the car at t = 1.5 s is at -8.1625 meters

4 0

3 years ago

What is the kinetic energy of a 1.40 kg discus with a speed of 22.5 m/s?

Oksana_A [137]

Kinetic energy = (1/2) (mass) (speed)²

   = (1/2) (1.4 kg) (22.5 m/s)²

   = (0.7 kg) (506.25 m²/s² )

   = 354.375 kg-m²/s² = 354.375 joules .

This is just the kinetic energy associated with a 1.4-kg glob of
mass sailing through space at 22.5 m/s. In the case of a frisbee,
it's also spinning, and there's some additional kinetic energy stored
in the spin.

8 0

3 years ago

Which of the following is not used in calculating acceleration?

Leno4ka [110]

Answer:

l think C am not pretty show

4 0

2 years ago