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g100num [7]
2 years ago
13

A piece of a metal alloy with a mass of 114 g was placed into a graduated cylinder that contained 25.0 mL of water, raising the

water level to 42.5 mL. What is the density of the metal?
Physics
1 answer:
Minchanka [31]2 years ago
8 0

Answer:

6.51 g/c.c

Explanation:

mass, m = 114 g

initial volume, V1 = 25 mL

final volume, V2 = 42.5 mL

Volume of the metal piece, V = V2 - V1 = 42.5 - 25 = 17.5 mL

1 mL = 1 c.c

So, Volume of metal, V = 17.5 c.c.

Let the density of the metal is d.

density = mass / volume

d = 114 / 17.5 = 6.51 g/c.c

Thus, the density of metal is 6.51 g/c.c.

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An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass
solniwko [45]

Answer:

A) The new amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please find the attached files for the solution

4 0
2 years ago
A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel
Vikki [24]

acceleration of rocket is given here as

a_y = 2.60* t

now we know that

\frac{dv}{dt} = 2.60t

now integrating both sides

\int dv = \int 2.60t dt

v = 2.60\frac{t^2}{2}

v = 1.30 t^2

here since its given that rocket will accelerate for t = 10 s

so here we have

v = 1.30 * 10^2

v = 130 m/s

so after t = 10 s the speed of rocket will be 130 m/s upwards

5 0
2 years ago
A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement? 4 miles 16 miles 20 miles 2
Ostrovityanka [42]
For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
 We have then
 d ^ 2 = 16 ^ 2 + 12 ^ 2

 From here, we clear the value of d.
 We have then:
 d =  \sqrt{16 ^ 2 + 12 ^ 2} 

 Rewriting:
 d = \sqrt{256 + 144}
 d = \sqrt{400}
 d = 20 miles

 Answer:
 
The magnitude of the car's displacement is:
 
d = 20 miles
7 0
2 years ago
Read 2 more answers
A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the
lions [1.4K]

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

3 0
2 years ago
for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha
VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

R=\dfrac{u^2\ \sin2\theta}{g}

Here,

u is the speed of launch of projectile

Here, \theta=30^{\circ}

We need to find the other launch angle when the projectile have the same range, such that,

\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}

\sin(60)=\sin2\alpha

\dfrac{\sqrt3}{2}=\sin2\alpha

\alpha =30^{\circ}

So, the other angle is 30 degrees. Hence, this is the required solution.

3 0
2 years ago
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