I really need help with this question someone plz help !

A block of mass 12.2 kg is sliding at an initial velocity of 3.9 m/s in the positive x-direction. The surface has a coefficient

Studentka2010 [4]

Answer:

Explanation:

a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force

R = mg where m is mass of the block

Force of friction F = μ x mg

= .173 x 12.2 x 9.8

= 20.68 N

b ) Only force of friction is acting on the body so

deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²

acceleration = - 1.7 m /s²

c )

v² = u² - 2 a s

v = 0 , u = 3.9 m /s

a = 1.7 m /s

0 = 3.9² - 2 x 1.7 x s

s = 4.47 m

5 0

3 years ago

According to a college survey, 22% of all students work full time. find the mean for the number of 3) students who work full tim

attashe74 [19]

Assuming that the students worldwide are being considered, because of the extremely large population, this can be considered as a binomial distribution. A normal distribution is used most usually as a fair approximation of the binomial. The mean is the expectation, therefore:
E[x] = np = (16)(0.22) = 3.52
μ = 3.52

3 0

3 years ago

Circuit diagram showing three resistors in series with an ammeter,a cell and a switch

docker41 [41]

Answer:

the mayflower

Explanation:

4 0

3 years ago

A 10 kg bear is running to the right at 10 m/s. He crashes into another 10 kg bear running to the left at -5 m/s. They collide a

faust18 [17]

Answer:

Total momentum is conserved

Explanation:

Conservation of Momentum

The momentum is a physical magnitude that measures the product of the object's velocity by its mass. The total momentum of a system is the sum of all its components' individual momentums. The two-bear system starts with a total moment of

$p=m_1v_1+m_2v_2=(10)(10)+(10)(-50)=50\ kg.m/s$

When both bears stick together, the total mass is 20 kg, and the new momentum is

$p'=(20)(2.5)=50 \ kg.m/s$

We have assumed both bears move to the right after the collision. In this situation, the total momentum is conserved

7 0

4 years ago

About 65 million years ago an asteroid struck Earth in the area of the Yucatán Peninsula and wiped out the dinosaurs and many ot

9966 [12]

Answer:

$2.44156\times 10^{13}\ m^3$

29010.53917 m

Explanation:

$\rho$ = Density of asteroid = 2 g/cm³

V = Volume

d = Diameter = 10 km

r = Radius = $\dfrac{d}{2}=\dfrac{10}{2}=5\ km$

v = Velocity = 11 km/s

$H_v$ = Heat vaporization of water = $2.26\times 10^6\ J/kg$

$\Delta T$ = Change in temperature = 100-20

Mass is given by

$m=\rho V\\\Rightarrow m=\rho\dfrac{4}{3}\pi r^3\\\Rightarrow m=2000\dfrac{4}{3}\times \pi\times 5000^3\\\Rightarrow m=1.0472\times 10^{15}\ kg$

The kinetic energy is

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1.0472\times 10^{15}\times 11000^2\\\Rightarrow K=6.33556\times 10^{22}\ J$

Heat is given by

$Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg$

Mass of water is $2.44156\times 10^{16}\ kg$

Volume is $\dfrac{2.44156\times 10^{16}}{10^3}=2.44156\times 10^{13}\ m^3$

Amount of water is $2.44156\times 10^{13}\ m^3$

If it were a cube

$h=V^{\dfrac{1}{3}}\\\Rightarrow h=(2.44156\times 10^{13})^{\dfrac{1}{3}}\\\Rightarrow h=29010.53917\ m$

The height of the water would be 29010.53917 m

4 0

3 years ago