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4 years ago

9

It's important to match your exercise shoes with the type of exercise in which you will be participating, Please select the best

answer from the choices provided OT

2 answers:

Feliz [49]4 years ago

8 0

Answer:

true

Explanation:

saveliy_v [14]4 years ago

6 0

Answer:

the answer is true

Explanation:

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An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the

garri49 [273]

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = $\dfrac{150}{45 cos 50^0}$

t = 5.19 s

a) height of arrow

$s = u t +\dfrac{1}{2}gt^2$

$s = v sin \theta \times t+\dfrac{1}{2}gt^2$

$s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2$

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = $\dfrac{u}{g}=\dfrac{30.29}{9.81}$

= 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s

5 0

3 years ago

To a stationary observer, a bus moves north with a speed of 10 m/s. A man inside walks toward the back of the bus with a speed o

Rudiy27

9.6m/s - apex .........................

3 0

3 years ago

Read 2 more answers

The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol

Lesechka [4]

Answer:

$\Delta V = 0.053 A$

Explanation:

Electric field in a given region is given by equation

$E = \frac{A}{r^4}$

as we know the relation between electric field and potential difference is given as

$\Delta V = -\int E. dr$

so here we have

$\Delta V = - \int (\frac{A}{r^4}) .dr$

$\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}$

here we know that

$r_1 = 1.71 m$ and $r_2 = 2.89 m$

so we will have

$\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})$

so we will have

$\Delta V = 0.053 A$

8 0

3 years ago

Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and

tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

&

q_x = -k*(dT/dx)

<u>Case (1) </u>

dT/dx= (-20-50)/0.35==> -280 K/m

q_x =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (3)

q_x =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0

3 years ago

Mechanical energy is conserved in the presence of which of the following types of forces?magnetic

Tju [1.3M]

<h2>Answer: electrostatic and gravitational force </h2><h2 />

Mechanical energy remains constant (conserved) if only <u>conservative forces</u> act on the particles.

In this sense, the following forces are conservative:

-Gravitational

-Elastic

-Electrostatics

While the Friction Force and the Magnetic Force are not conservative.

According to this, mechanical energy is conserved in the presence of electrostatic and gravitational forces.

7 0

4 years ago