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3 years ago

Question 2 (1 point)

Physics

1 answer:

Tju [1.3M]3 years ago

7 0

Answer:

I know someone anwsered but it would be 400M

Explanation:

i initial velocity (u)=10m/s

acceleration (a)=0

time taken (t) =40s

then distance (s)=u t +1/2 a t^2

s= u t +0 (as a is 0)

s= 10 x 40

s= 400M

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A 4.00 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0 degrees with t

steposvetlana [31]

Answer:

0.35

Explanation:

According to Newton's second law;

\sum Fx = ma

Fm - Ff =ma

Fm is the moving force = Wsin theta

Fm = 4(9.8)sin55

Fm = 32.1N

Ff is the frictional force = nmgcos theta

Ff = n(4)(9.8)cos55

Ff = 22.48n

Acceleration a = 6.0m/s²

Substitute the given values into the formula and get the coefficient of friction

32.11-23.48n = 4(6)

32.11-24= 23.48n

8.11 = 23.48

n = 8.11/23.48

n = 0.35

Hence the coefficient of friction is 0.35

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3 years ago

Mike stands on a scale in an elevator. If the elevator is accelerating upwards with 4.9 m/s2, the scale reading is ____ times Mi

Bas_tet [7]

Answer:

F - M a force exerted by scales on student

M a = M (9.8 + 4.9) m/s^2 upwards chosen as positive

a = 1.5 g net acceleration of student due to force of scales

W =M g weight of student (actual weight)

Wapp = M 1.5 * g apparent weight (on scales) of student

7 0

3 years ago

A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swin

Goshia [24]

Answer:

(A) 0.63 J

(B) 0.15 m

Explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

angular speed (ω) = 4 rad/s

To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + m $h^{2}$

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I = $(\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2}$ )^{2}[/tex]

I = $(\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2}$ )^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5I $ω^{2}$

= 0.5 x 0.07875 x $4^{2}$ = 0.63 J 0.15 m

(B) from the conservation of energy

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

Ki + Ui = Kf + Uf

at the maximum height velocity = 0 therefore final kinetic energy = 0

Ki + Ui = Uf

Ki = Uf - Ui

Ki = mg(H-h)

where (H-h) = rise in the center of mass

0.63 = 0.42 x 9.8 x (H-h)

(H-h) = 0.15 m

6 0

3 years ago

A factory emits pollutants at a rate of 25 g/s. The factory is located between two mountain ranges resulting in an effective val

Lostsunrise [7]

Answer:

$1.25\ \mu\text{g/m}^3$

Explanation:

v = Velocity of the breeze = 4 m/s

w = Width of the valley = 5000 m

h = Height of the valley = 1000 m

Volumetric flow rate is given by

$\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}$

$\dot{m}$ = Mass flow rate of pollutant = 25 g/s = $25\times 10^6\ \mu\text{g/s}$

Concentration is given by

$C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3$

The steady state concentration of pollutants in the valley, is $1.25\ \mu\text{g/m}^3$ .

6 0

3 years ago

The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t

maxonik [38]

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

The work done by the resistance force.

Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s$

Work done is equal to the change in kinetic energy. It can be given as follows :

$W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J$

So, the required work done is 32.99 J.

3 0

3 years ago