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iragen [17]
2 years ago
9

You drop your keys in a high-speed elevator going up at a constant speed. Part APart complete Do the keys accelerate faster towa

rd the elevator floor than they would if the elevator were not moving? Do the keys accelerate faster toward the elevator floor than they would if the elevator were not moving? Yes No Previous Answers Correct Part B Do the keys accelerate faster toward the elevator floor than they would if the elevator were accelerating downward? Do the keys accelerate faster toward the elevator floor than they would if the elevator were accelerating downward? No Yes
Physics
1 answer:
anzhelika [568]2 years ago
5 0

Answer:

Explained

Explanation:

a) No, the keys were initially moving upward in the elevator only effects the initial velocity of the key and not the rate of change of velocity that is acceleration. So, the keys accelerate with the same acceleration as before.

b)Yes, keys will accelerate towards the floor faster if it is a constant speed than it is moving downward because if the elevator is accelerating downward, the downward change in velocity of the keys is at least partially matched by a downward change in the velocity of the of the elevator.

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The net force on a boat causes it to accelerate at 1.55 m/s2. The mass of the boat is 215 kg. The same net force causes another
jekas [21]

Answer:

2666 kg

0.11567 m/s²

Explanation:

m = Mass of boat

a = Acceleration of boat

From Newton's second law

Force

F=ma\\\Rightarrow F=215\times 1.55\\\Rightarrow F=333.25\ N

Force on the first boat is 333.25 N

F=ma\\\Rightarrow m=\frac{F}{a}\\\Rightarrow m=\frac{333.25}{0.125}\\\Rightarrow m=2666\ kg

Hence, mass of the second boat is 2666 kg

Combined mass = 2666+215 = 2881 kg

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{333.25}{2881}\\\Rightarrow a=0.11567\ m/s^2

The acceleration on the combined mass is 0.11567 m/s²

6 0
2 years ago
A wire of 1mm diameter and 1m long fixed at one end is stretched by 0.01mm when a lend of 10 kg is attached to its free end.calc
Otrada [13]

Answer:

E = 1.25×10¹³ N/m²

Explanation:

Young's modulus is defined as:

E = stress / strain

E = (F / A) / (dL / L)

E = (F L) / (A dL)

Given:

F = 10 kg × 9.8 m/s² = 98 N

L = 1 m

dL = 10⁻⁵ m

A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²

Solve:

E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)

E = 1.25×10¹³ N/m²

Round as needed.

5 0
2 years ago
At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c
fiasKO [112]

Answer:

497.00977 N

3742514.97005

Explanation:

\rho = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N

The drag force on the dolphin's nose is 497.00977 N

at 20°C

\mu = Dynamic viscosity = 1.002\times 10^{-3}\ Pas

Reynold's Number

Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005

The Reynolds number is 3742514.97005

8 0
2 years ago
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block
melomori [17]

This question is incomplete, the complete question is;

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.

How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°

Answer:

the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J

Explanation:

Given that;

m = 9.9 kg

h = 4.9 m

d = 5 m

μ = 0.3

θ = 36.87°

Now from conservation of energy, the energy is;

Et = mgh

we substitute

Et = 9.9 × 9.8 × 4.9

= 475.398 J

Also the loss of energy i

E_loss = (umg cosθ) d

we substitute

E_loss  = 0.3 × 9.9 × 9.8 × cos36.87°  × 5

= 116.423 J

so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be

E = Et - E_loss

E = 475.398 J - 116.423 J

E = 358.975 J

5 0
2 years ago
Which combination of initial horizontal velocity, (vh) and initial vertical velocity, (vv) results in the greatest horizontal ra
Naya [18.7K]

If an object is projected with vertical speed given as

v_v

now the time of flight of the object that time in which it comes back on ground

\Delta y = v_v t + \frac{1}{2}(-g)t^2

now here we will have

0 = v_v t - \frac{1}{2}gt^2

t = \frac{2v_v}{g}

now the range of projectile is given as

R = horizontal\: speed \times time

R = v_h(\frac{2v_v}{g}

now here we know that

v_v = v_0 sin\theta

v_h = v_0 cos\theta

now the range is given as

R = \frac{2(vsin\theta)(vcos\theta)}{g}

R = \frac{v^2sin2\theta}{g}

now in order to have maximum range we can say

sin2\theta = 1

2\theta = 90^0

so we will have

\theta = 45 ^0

so now we can say

v_v = v_h = \frac{v_0}{\sqrt2}

so both speed must be same to have maximum horizontal range

8 0
2 years ago
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