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PtichkaEL [24]
3 years ago
13

Physics questions please help me

Physics
1 answer:
Ahat [919]3 years ago
5 0
Hi,Find  answers from Task 5

1.(X+4)+(X)+(X+4)+(X)=50cm

4x+8=50cm

4x=42

X=10.5cm

Length=10.5+4=14.5cm

Width=10.5cm

Area= length × width=(10.5/100) × (14.5/100) =0.0152m2

2. Volume of a sphere= 4/3 ×π×r³

4/3 ×π×r³=3.2×10^-6 m³

r³=3.2×10^-6 m³/1.33×π

r³=7.64134761e-7

r=0.00914m

Surface area of the blood drop= 4πr²

=4×3.142×0.00914×0.00914=0.00105m²

3.

Equation of an ideal gas     = PV =n RT  

Equation for pressure, = P= n RT/V

Equation for the volume of an ideal gas= V= n RT/P

If the volume of gas doubles ,V(new)=  2n RT/P

Equation for temperature of an ideal gas, T = PV/n R

If temperature of gas triples, T (new)= 3PV/n R

New Equation for Pressure, = n× R× (3PV/n R)/(2n RT/P)

Pressure factor increase= P(new)/P(old)  ={ n× R× (3PV/n R)/(2n RT/P)}/{ n RT/V}

=3PV²/2n RT
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If you increase the mass of an object and want to move an object a specific distance, what do you need to do
Alexxandr [17]

If you increase the mass of an object and want to move an object a specific distance, then you need to do extra work than the earlier

<h3>What is work done?</h3>

The total amount of energy transferred when a force is applied to move an object through some distance

Work Done = Force * Displacement

For example, let us suppose a force of 10 N is used to displace an object by a displacement of 5 m then the work done on the object can be calculated by the above-mentioned formula

work done = 10 N ×5 m

                 =50 N m

Thus, when an object's mass is increased and it is desired to move it a certain distance, more work must be done than previously.

Learn more about work done from here

brainly.com/question/13662169

#SPJ1

3 0
2 years ago
What is the electric force acting between two charges of -0.0045 C and -0.0025 C that are 0.0060 m apart? Use Fe=kq1q2/r^2 and k
oksano4ka [1.4K]

Answer:

D. 2.8 × 10⁹ N

Explanation:

The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.

Fe= k Q₁Q₂/r²

Q₁= -0.0045 C

Q₂= -0.0025 C

r= 0.0060 m

k= 9.00 × 10 ⁹ Nm²/C²

Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²

=2.8 × 10⁹ N

4 0
4 years ago
Read 2 more answers
If the velocity of a car is 45 km/h west, how far can it travel in 0.5 hours?
Anarel [89]
45km/h * 0.5h= 22.5km

The car can travel 22.5km in 0.5 hours
6 0
3 years ago
Read 2 more answers
An 3.7 lb hammer head, traveling at 5.8 ft/s strikes a nail and is brought to a stop in 0.00068 s. The acceleration of gravity i
CaHeK987 [17]

Answer:

31677.2 lb

Explanation:

mass of hammer (m) = 3.7 lb

initial velocity (u) = 5.8 ft/s

final velocity (v) = 0

time (t) = 0.00068 s

acceleration due to gravity (g) 32 ft/s^{2}

force = m x ( a + g )

where

  • m is the mass = 3.7 lb
  • g is the acceleration due to gravity = 32 ft/s^{2}
  • a is the acceleration of the hammer

       from v = u + at

       a = (v-u)/ t

       a = (0-5.8)/0.00068 = -8529.4 ( the negative sign showa the its decelerating)

we can substitute all required values into force= m x (a+g)

force = 3.7 x (8529.4 + 32) = 31677.2 lb

       

4 0
3 years ago
A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration
Rufina [12.5K]

To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

F = ma

Where,

m= Mass

a = Acceleration

At the same time the frictional force can be defined as,

F_f = \mu N

Where,

\mu = Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

m_h = 242 kg\\m_c = 224 kg\\\mu = 0.894\\

By condition of Balance the friction force must be equal to the total net force, that is to say

F_{net} = F_f

m_{total}a = \mu m_hg

(m_h+m_c)a = \mu*m_h*g

Re-arrange to find acceleration,

a= \frac{\mu*m_h*g}{(m_h+m_c)}

a = \frac{0.894*242*9.8}{(242+224)}

a = 4.54 m/s^2

Therefore the acceleration the horse can give is 4.54m/s^2

3 0
3 years ago
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