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SCORPION-xisa [38]
3 years ago
6

1. Density = 13.5 g/mL Mass = 151 grams Volume = What is the volume

Physics
2 answers:
maxonik [38]3 years ago
5 0

Answer:

density = mass /vol

13.5 =151/ vol

vol = 151/13.5 = 11.2 ml

stealth61 [152]3 years ago
4 0
Answer: Volume = 11.18mL
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plz help me with hw A bus of mass 1000 kg moving with a speed of 90km/hr stops after 6 sec by applying brakes then calculate the
Lelechka [254]

Answer:

Mass, M = 1000 kg

Speed, v = 90 km/h = 25 m/s

time, t = 6 sec.

Distance:

{ \tt{distance =  speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}

Force:

{ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion :  \\ { \bf{s = ut +  \frac{1}{2}  {at}^{2} }} \\  \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times  {6}^{2} ) }} \\  \\ { \tt{acceleration = 8.33 \:  {ms}^{ - 2} }} \\  \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}

5 0
3 years ago
An athlete stretches a spring an extra 28.6 cm beyond its initial length. how much energy has he transferred to the spring, if t
g100num [7]
PE = 0.5 × k × x²

PE potential Energy
k spring constant
x stretch/compression of the spring
6 0
3 years ago
Energy from the Sun arrives at the top of the Earth’s atmosphere with an intensity of 1.30kW/m21.30⁢kW/m2. How long does it take
vivado [14]

Answer:

T=1.384×10⁶seconds

Explanation:

Given data

p (Intensity)=1.30 kw/m²

E (Energy)=1.8×10⁹ J

A (Area)=1.00 m²

T (Time required)=?

Solution

E=PT   ................eq(i)

where E is energy

P is radiation power

T is time

Radiating Power is given as

P=pA

Where p is intensity

A is Area

Put P=pA in eq(i) we get

E=pAT

T=E/pA

T=\frac{(1.8*10^{9}}{(1.30*10^{3} )*(1.00)}  )\\T=1.38*10^{6} seconds

3 0
3 years ago
You have two identical boxes. You place skittles in one and the snicker bars in the other. Which box can hold more pieces of can
vichka [17]
The skittles because they are small so you can fit more in
3 0
3 years ago
A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

4 0
3 years ago
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