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2 years ago

an ice skater is moving at a speed of 2 mi/hr when she slams into a wall 0.05 hours later. what is her acceleration?

Physics

2 answers:

Firdavs [7]2 years ago

7 0

Answer:

0.00497m/s²

Explanation:

Acceleration is defined as the change in velocity of a body with respect to time. Mathematically,

Acceleration = change in velocity/time

Given velocity = 2miles/hour

Converting mi/hr to m/s,

1mi/hr = 0.447m/s

2mi/hr = 2×0.447

= 0.894m/s

Time taken to slam into a wall = 0.05hours

Converting this to seconds will give;

0.05×60×60

= 180seconds

Acceleration = 0.894/180

Acceleration = 0.00497m/s²

nikklg [1K]2 years ago

6 0

$v_{f} = v_{0}+at$
at $v_{f}=0$
$v_{0}=at$
$-2=0.05a$
$a= \frac{-2}{0.05} =-40mi/hr$

Hope this helps.

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Two trucks with equal mass are attracted to each other with a gravitational force of 5.3 x 10 -4 N. The trucks are separated by

HACTEHA [7]

Answer:

7327 kg or 7.3 tons

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408*10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M = M_1 = M_2$ is the masses of the 2 objects. and R = 2.6m is the distance between them.

$F_G = 5.3*10^{-4}N$ is the attraction force.

$5.3*10^{-4} = 6.67408*10^{-11}\frac{M^2}{2.6^2}$

$7941265 = \frac{M^2}{2.6^2}$

$M^2 = 53682948.76$

$M = \sqrt{53682948.76} \approx 7327 kg$ or 7.3 tons

3 0

2 years ago

A skier is moving down a snowy hill with an acceleration of 0.40 m/s2. The angle of the slope is 5.0∘ to the horizontal. What is

kirill115 [55]

Answer:

1.25377 m/s²

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of friction

$\theta$ = Slope

From Newton's second law

$mgsin\theta-f=ma\\\Rightarrow mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow \mu=\frac{gsin\theta-a}{gcos\theta}\\\Rightarrow \mu=\frac{9.81\times sin5-0.4}{9.81\times cos5}\\\Rightarrow \mu=0.04655$

Applying $\mu$ to the above equation and $\theta=10^{\circ}$

$mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow a=gsin\theta-\mu gcos\theta\\\Rightarrow a=9.81\times sin10-0.04655\times 9.81\times cos10\\\Rightarrow a=1.25377\ m/s^2$

The acceleration of the same skier when she is moving down a hill is 1.25377 m/s²

3 0

2 years ago

A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

2 years ago

A cyclist moving towards right with an acceleration of 4m/s² at t = 0 he has travelled 5 m moving towards the right at 15 m/s wh

ira [324]

Answer:

$x=2t^2+15t+5$

Explanation:

$x=\frac{1}{2}at^2+v_0t+x_0$

4 0

2 years ago

Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous

Orlov [11]

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is Av with A the cross section area and v the velocity,

so,

$A_1V_1=A_2V_2$

$A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}$

the diameter decreases 86% so

$d_2 = 0.86d_1$

$v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}$

Thus, speed in smaller section is 48.6 m/s

3 0

2 years ago