Ask question

Ask question

All categories

3 years ago

6

The hypotenuse of a right triangle is 29.0 centimeters. The length of one of its legs is 20.0 centimeters. What is the length of

the other leg?

2 answers:

kirza4 [7]3 years ago

7 0

A = 21
Hope this Helps!

SVEN [57.7K]3 years ago

3 0

C2 = a2 + b2
29^2 = a^2 + 20^2
841 = a^2 + 400
441 = a^2
a = 21

You might be interested in

boris the bear is lumbering along at a constant rate of 4.0 m/s. if he walks for 600.0 seconds, how far will he travel?

irina1246 [14]

Answer: 2400m

Explanation: 2400m because 600 times 4 equals 2400

6 0

3 years ago

Bodies A and B have equal mass. Body B is initially at rest. Body A collides with body B in a one-dimensional elastic collision.

jek_recluse [69]

According to the statement " Collision <span>between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision."
The best answer is :
Option A " </span><span>BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A "</span>

4 0

3 years ago

Read 2 more answers

A combination lock has a 1.3-cm-diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you twist

galina1969 [7]

Answer:

0.04225 Nm

Explanation:

N = Force applied = 5 N

$\mu$ = Coefficient of static friction = 0.65

d = Diameter of knob = 1.3 cm

r = Radius of knob = $\frac{d}{2}=\frac{1.3}{2}=0.65\ cm$

Force is given by

$F=N\mu\\\Rightarrow F=5\times 0.65\\\Rightarrow F=3.25\ N$

When we multiply force and radius we get torque

Torque on thumb

$\tau_t=F\times r\\\Rightarrow \tau_t=3.25\times 0.0065\\\Rightarrow \tau_t=0.021125\ Nm$

Torque on forefinger

$\tau_f=F\times r\\\Rightarrow \tau_f=3.25\times 0.0065\\\Rightarrow \tau_f=0.021125\ Nm$

The total torque is given by

$\tau=\tau_t+\tau_f\\\Rightarrow \tau=0.021125+0.021125\\\Rightarrow \tau=0.04225\ Nm$

The most torque that exerted on the knob is 0.04225 Nm

4 0

3 years ago

What describes a sound wave as it travels through a medium

Softa [21]

Sound waves in air (and any fluid medium) are longitudinal waves because particles of the medium through which the sound is transported vibrate parallel to the direction that the sound wave moves.

5 0

3 years ago

where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed

lesya [120]

Answer:

$E_r(6)=4.35614\ MPa$

Explanation:

$\epsilon$ = Strain = 0.49

$\sigma _0$ = 3.1 MPa

At t = Time = 32 s $\sigma$ = 0.41 MPa

$\tau$ = Time-independent constant

Stress relation with time

$\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)$

at t = 32 s

$0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s$

The time independent constant is 16.0787 s

$E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}$

At t = 6

$\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}$

From the first equation

$\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451$

$E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa$

$E_r(6)=4.35614\ MPa$

6 0

3 years ago