Ask question

Ask question

All categories

2 years ago

6

HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLE

ASE

1 answer:

Romashka [77]2 years ago

6 0

$A_x = 5.0$

$A_y = -6.3$

You might be interested in

A radio station broadcasts at a frequency of 600 kHz. Knowing that radio waves have a speed of 300 000 000 m/s, what is the wave

soldier1979 [14.2K]

Answer:

ccvtesgdujtdchgdrgggggggfrrrtyfaasdddfffghgdshh

6 0

3 years ago

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.

crimeas [40]

Answer:

Explanation:

Given:

length of ladder $r_L = 14m$

weight of ladder $F_L = 490N$

position of firefighter $r_F = 3.8m$

weight of firefighter $F_F = 820N$

angle of ladder $\alpha = 63$

Unknown:

force of the wall on the ladder $F_W$

force of friction on base of ladder $F_R$

normal force on base of ladder $F_N$

From the free body diagram of the sketch you get 3 equations:

$F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0$

Solving the equations gives:

$F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}$

a)

$F_R = 238N\\ F_N = 1310N$

b)

$F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3$

c) Using the result from b and solving for $r_F$

$\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m$

4 0

3 years ago

A brick sliding across a smooth floor has a coefficient of static friction of s=0.23 and a coefficient of kinetic friction of k=

pantera1 [17]

Answer:

0.029325

Explanation:

0.23x0.15=0.0345

0.0345x0.85=0.029325

4 0

1 year ago

Read 2 more answers

A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that

saveliy_v [14]

Answer:

The speed it reaches the bottom is

$v=6.51m/s$

Explanation:

Given: $m=5.0kg$ , $r=47cm\frac{1m}{100cm}=0.47m$

Using the conservation of energy theorem

$U_i=K_E+K_{ER}$

$m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2$

$v=r*w$ , $I=\frac{1}{2}*m*r^2$

$m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2$

$m*g*h=\frac{3}{4}*m*r^2*w^2$

$g*h=\frac{3}{4}*r^2*w^2$

Solve to w'

$w^2=\frac{4*g*h}{3*r^2}$

$h=x*sin(30)=6.5m*sin(30)=3.25m$

$w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}$

$w=27.74rad/s$

$v=27.74rad/s*0.235m=6.51m/s$

7 0

3 years ago

If the wavelength of a wave is quadrupied then what happens to the frequency of the wqve(Assume speed is constant)

dimulka [17.4K]

Answer:

What happens to the wavelength of a wave if you double the frequency?

If the frequency of a wave is increased, what happens to its wavelength? As the frequency increases, the wavelength decreases. 2. If the frequency is doubled, the wavelength is only half as long.

Explanation:

4 0

2 years ago