Can you answer one of the questions

Which formula can be used to find the angle of the resultant vector has

nata0808 [166]

The diagram shows components that have been added together to form Rx and Ry. Rx and Ry are the components of the resultant vector.
Which formula can be used to find the angle of the resultant vector?
the answer is C
C. tan0= Ry/Rx

5 0

2 years ago

Which of these is not a part of the<br> cardiovascular system?

lilavasa [31]

Answer:I don't see any

Explanation:

6 0

3 years ago

Read 2 more answers

A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic

Ilia_Sergeevich [38]

Answer:

$\mu_k=0.27$

Explanation:

According to the free body diagram, in this case, we have:

$\sum F_x:-F_k=ma\\\sum F_y:N=mg$

Recall that the force of friction is given by:

$F_k=\mu_k N$

Replacing and solving for the coefficient of kinetic friction:

$-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}$

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

$a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}$

Finally, we calculate $\mu_k$ :

$\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27$

4 0

3 years ago

Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that

lawyer [7]

As we know by the first law of thermodynamics

$Q = \Delta U + W$

here we know that

Q = heat given to the system

$\Delta U = U_f - U_i$

W = work done by the system

now here we can say

$\Delta U = 180 - 78 = 102 J$

$W = 64 J$

now we can say that heat will be given as

$Q = 64 + 102 = 166 J$

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy

So best describe Jin's Error is

<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>

8 0

3 years ago

A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas

Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

(c) the maximum acceleration of the object, a_max = 15.4 m/s²

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

$V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s$

(c) the maximum acceleration of the object

$a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2$

6 0

2 years ago