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Pachacha [2.7K]
1 year ago
7

Anna is sitting in a moving cart and throws a ball straight up. Theoretically, the ball should land in the cart, but it lands on

the ground a little before the cart. What is the reason for the difference in the ranges of the ball and the cart? There is no external horizontal force acting on the ball or the cart. There is no external vertical force acting on the ball or the cart. There is a difference in the external horizontal forces acting on the ball and the cart. There is a difference in the external vertical forces acting on the ball and the cart.
Physics
2 answers:
anastassius [24]1 year ago
8 0

The guys before me was almost right. But instead of it being vertical it is horizontal.

Temka [501]1 year ago
8 0

Answer:

There is a difference in the external horizontal forces acting on the ball and the cart.

Explanation:

When ball is thrown upwards from a moving cart then in that case the acceleration of the ball in vertical direction is due to gravity so it will raised up and again come back to ground.

While during this vertical motion it will also move horizontally with initial speed of cart at which it was moving at the moment when ball is thrown.

Now if ball is not landed on the cart again then we can say that both cart and ball are not accelerated with same acceleration in horizontal direction due to which they both will cover different distances in horizontal direction for the same time interval

So we can say that correct answer will be

There is a difference in the external horizontal forces acting on the ball and the cart.

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A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
ZanzabumX [31]

Answer:

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b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0
1 year ago
A 1700kg rhino charges at a speed of 50.0km/h. What is the magnitude of the average force needed to bring the rhino to a stop in
Ksivusya [100]
Force (N) = mass (kg) x velocity (m/s) / time (s)
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2 years ago
A 4 kg box is on a frictionless 35° slope and is connected via a massless string over a massless, frictionless pulley to a hangi
Anarel [89]

Answer:

(a) 19.62 N

(b) Box moves down the slope

(c) 24.43 N

Explanation:

(a)  

2 Kg box  causes tension

T=mgwhere m is mass, g is gravitational force taken as 9.81T=2*9.81 =19.62 N  (b)  Block mass of 4 Kg  [tex]T'-mg sin \theta=0 hence T'=mg sin \theta where m is mass and g is gravitational force  

T'=4*9.81 sin 35= 22.5071 N  

Since T' is greater than mg sin\theta , then the box moves down the slope  

(c)  

Acceleration a= \frac {forward   force-backward   force}{Total mass}= \frac {mg sin \theta -mg}{m1 + m2}  

a= \frac {22.51-19.62}{2+4}=0.48

When moving, the box will exert force T"= mgsin \theta + ma  

T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N

7 0
2 years ago
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Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerate
____ [38]

Answer:

a) Velocity = 4.2m/s

b) Acceleration = 2.94m/s^2

c) Force exerted on the floor= 1401.4×10^3N

Explanation:

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V= 4.2m/s

b) Vf2= V^2+2ay2

a= 4.2^2 - 0/2×3

a= 17.64/6= 2.94m/s^2

c) Newton's 2nd law indicates:

Fnet= F - mg=ma

F= m(g+a)

F=110(9.8+2.94)

F=110×12.94

F= 1401.4N

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1 year ago
Backpackers often use canisters of white gas to fuel a cooking stove's burner. If one canister contains 1.45 L of white gas, and
lorasvet [3.4K]

Answer: One canister contains 1.03 Kg of fuel,

Explanation:

The density is defined as the relation between the mass and the volume p=\frac{m}{v}

First of all you need to have the same units for volume and density, so:

For the volume,

1.45 L * \frac{1.10*10^{-3}m^{3}  }{1 L} = 1.45 *10^{-3} m^{3}

For the density,

p = 0.710\frac{g}{cm^{3} } *\frac{1cm^{3} }{1*10^{-6}m^{3}  }*\frac{1 Kg}{1*10^{3}g } = 710\frac{Kg}{m^{3} }

From the density equation we have m=p*v, so:

710\frac{Kg}{m^{3} } *1.45*10^{-3} m^{3} = 1.03Kg

4 0
2 years ago
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