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Dmitriy789 [7]
3 years ago
8

A 2-column table with 5 rows. The first column titled metal has entries aluminum, cork, iron, lead, wax. The second column title

d Density (grams per cubic centimeter) has entries 2.64, 0.24, 7.50, 11.34, 0.72. If the density of water is 1.0 g/cm3, which of these materials would float in water, based on their densities? Check all that apply. aluminum cork iron lead wax
Physics
2 answers:
astraxan [27]3 years ago
9 0

Answer:Cork and wax

Explanation:

ruslelena [56]3 years ago
8 1

Answer: Iron

Explanation:

I did it on edge 2020 and got it correct tytyty

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A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t
goldfiish [28.3K]

Answer:

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

y=y_{0}+uSin\theta t-1/2 gt^{2}

y=2+20Sin5 t-4.9t^{2}

As it hits the ground in time t, so put y = 0

0=2+1.74 t-4.9t^{2}

4.9t^{2}-1.74t-2=0

t= \frac{1.74\pm\sqrt{1.74^{2}+4\times\2\times4.9}}{9.8}

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X =  16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

y'=y_{0}+uSin\theta t'-1/2 gt'^{2}

y'=2+20Sin5 t-4.9\times0.35^{2}

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

3 0
2 years ago
A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (
Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants (v_{oy} = 0) zero, so let's use the equation

     y =v_{oy} t -1/2 g t²

     y= - ½ g t²

     t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

     x = vox t

     x = v₀ₓ √2y/g

c) Speeds before touching the ground

     vₓ = vox = constant

     v_{y} = v_{oy} - gt

     v_{y} = 0 - g √2y/g

    v_{y}  = - √2gy

    tan θ = Vy / vx

    θ = tan⁻¹ (vy / vx)

    θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0
3 years ago
Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 81 mN when sepa
Gekata [30.6K]

Answer:

Explanation:

Check the attachment for solution

8 0
3 years ago
A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you
soldi70 [24.7K]

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}       (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s

The interaction time to avoid that the water balloon breaks is 0.029s

5 0
3 years ago
PLEASE HELP ME I NEED THIS ASAP!!!!
Zina [86]

Answer:

loud bangs

Explanation:

the pots for cooking fell

3 0
3 years ago
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