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3 years ago

11

Unscramble the bolded letters to guess the 6 letter word code. Did you get the poses or exercises correct? MOUNTAIN POSE TRIANGL

E POSE MOUNTAIN CLIMBERS PLANK TREE POSE JUMPING JACKS PASSWORD Think a type of flexibility ??? Hold that stretch

1 answer:

nikitadnepr [17]3 years ago

8 0

Explanation:

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Acceleration question plz help

andrezito [222]

For #5 It's helpful to draw a free body diagram so you know which way the forces are acting on the block.

the weight mg is acting downwards, and you need to find the vertical and horizontal components of mg using sin and cosine. so do 15x9.8xsin40 which is the force. Assuming no friction, this is the only force acting on the block, as the forces on the vertical plane cancel out i.e the normal force and weight of the block.

after, just do F=ma And since you know F and m, solve for a.

3 0

4 years ago

A block with a mass of 31.8 kg is pushed on a frictionless

OlgaM077 [116]

Answer: Total work done on the block is 3670.5 Joules.

Step by step:

Work done:

$W = F.d.\cos\theta$

With F the force, d the displacement, and theta the angle of action (which is 0 since the block is pushed along the direction of displacement, and cos 0 = 1)

$W = F.d$

Given:

F = 75 N

m = 31.8 kg

Final velocity $v_f = 15.3 \frac{m}{s}$

In order to calculate the Work we need to determine the displacement, or distance the block travels. We can use the information about F and m to first figure out the acceleration:

$F = ma\\\implies a=\frac{F}{m}=\frac{75 N }{31.8 kg}\approx 2.36 \frac{m}{s^2}\\$

Now we can determine the displacement from the following formula:

$d = \frac{1}{2}a^2+v_0t+d_0$

Here, the initial displacement is 0 and initial velocity is also 0 (at rest):

$d = \frac{1}{2}at^2\\$

Now we still have "t" as unknown. But we are given one more bit of information from which this can be determined:

$v_f = a\cdot t_f\\\implies t_f = \frac{v_f}{a} = \frac{15.2 \frac{m}{s}}{2.36 \frac{m}{s^2}}\approx 6.44 s$

(using vf as final velocity, and tf as final time)

So it takes about 6.44 seconds for the block to move. This allows us to finally calculate the displacement:

$d = \frac{1}{2}at^2=\frac{1}{2}2.36 \frac{m}{s^2}\cdot 6.44^2 s^2 \approx 48.94 m$

and the corresponding work:

$W = F\cdot d=75 N\cdot 48.94 m =3670.5J$

7 0

3 years ago

What is the ideal mechanical advantage of the pulley system shown in this figure?

olchik [2.2K]

Answer:

v

Explanation:

vv

4 0

3 years ago

A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition

Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So $mg=Kx$

$K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m$

Now we know that $\omega =\sqrt{\frac{K}{m}}$

So $\frac{2\pi }{T} =\sqrt{\frac{K}{m}}$

$\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}$

$T=2.1371sec$

8 0

3 years ago

2. What is the speed of a car that travels 73.4 kilometers (km) in 5 hours?

maksim [4K]

Speed =dist./time
=73.4/5
=14.68 km/hr

7 0

3 years ago

Read 2 more answers