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2 years ago

A woman is applying 300N/m2 of pressure on to door with her hand. Her hand has area of 0.02m2. Work out the force being applied

Physics

1 answer:

never [62]2 years ago

8 0

Answer:

Explanation:

Given parameters:

Pressure applied by the woman = 300N/m²

Area = 0.02m²

Unknown:

Force applied = ?

Solution:

Pressure is the force per unit area on a body

Pressure = $\frac{force}{area}$

Force = Pressure x area

Force = 300 x 0.02 = 6N

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Q 32.35: The isotope 235U decays by alpha emission with a half-life of 7.0 x 108 y. It also decays (rarely) by spontaneous fissi

Julli [10]

Answer:

rate of fission =5.89*10^3 1\Year

Explanation:

we know that

rate of fission is given as $rate of fission = \frac{0.69}{(T_{1/2})_{fission}} *\frac{ Mass* Avogardo\ number}{Molar\ mass}$

$(T_{1/2})_{fission} = 3*10^{17} y$

mass = 1.0 g

avogardo number = 6.02*10^23

molar mass of isotopes 235U =235

Putting all value to get rate of emission

rate of fission $= \frac{0.69}{3*10^{17}} *\frac{1.0*6.02*10^{23}}{235}$

rate of fission =5.89*10^3 1\Year

5 0

2 years ago

A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second

Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

$v_x = 15 m/s$

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

$d= v_x t =(15 m/s)(5 s)=75 m$

5 0

2 years ago

Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m

Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

$m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s$

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

$V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

Substituting,

$V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$V_{cm} = 1.034m/s$

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where here $v_f$ is the velocity after the collision.

$v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

$v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$v_f = 1.034m/s$

8 0

2 years ago

Ronald likes to use his erector set more than anything else.

Rashid [163]

C: Foreclosure. People in identity foreclosure have committed to an identity too soon. Often they have simply adopted the identity of a parent, close relative or respected friend.

5 0

2 years ago

Read 2 more answers

Using the mass and radius of the Sun given under the "Vital Statistics" section of Chapter 10 in your textbook, calculate the Su

atroni [7]

Answer:

The density of the sun is 4434kg/m³

This was found by dividing the mass (1.989 ×10³⁰kg) by the volume (4.486×10²⁶ m³) which was calculated using V = 4/3×pi ×r³

Explanation:

See attachment below.