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3 years ago

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3. If you had parental units of RR and Rr (R being dominant round and r being recessive

1 answer:

yanalaym [24]3 years ago

7 0

C all round

Explanation:

because the outcome of a the traits would be RR, RR, Rr, Rr so it would be all round if it was Rr and Rr than there would have RR, Rr, Rr, and rr and that would have been a

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Nataly [62]

Your answer is C)

a)t=2.78 sec

b)R=835.03 m

c)

Explanation:

Given that

h= 38 m

u=300 m/s

here given that

The finally y=0

So

t=2.78 sec

The horizontal distance,R

R= u x t

R=300 x 2.78

R=835.03 m

The vertical component of velocity before the strike

4 0

3 years ago

Friction is a force in which two objects __________.

balu736 [363]

The answer should be C) Slide against each other.

7 0

3 years ago

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Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on

leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

$q_1=50\ nC at y=6\ m$

$q_2=-80\ nC at x=-4\ m$

$q_3=70\ nC at y=-6\ m$

Electric Potential is given by

$V=\frac{kQ}{r}$

Distance of $q_1$ from $x=8\ m$

$d_1=\sqrt{6^2+8^2}$

$d_1=\sqrt{36+64}$

$d_1=10\ m$

similarly $d_2=8-(-4)$

$d_2=12\ m$

$d_3=\sqrt{(-6)^2+8^2}$

$d_3=\sqrt{36+64}$

$d_3=10\ m$

Potential at $x=8\ m$ is

$V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}$

$V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]$

$V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}$

$V_{net}=9\times 5.33$

$V_{net}=47.97\approx 48\ V$

5 0

3 years ago

A block is pulled along a horizontal surface at a constant speed by a force (14.1i 0 j 5.1k). The direction k is perpendicular t

Pepsi [2]

Answer:

$W =84.6\ Nm$

Explanation:

given,

F = 14.1 i + 0 j + 5.1 k

displacement = 6 m

Assuming block is moving in x- direction

we know,

dW = F dx

$\int dW = F\int dx$

$W = F\int_0^6 dx$

$W = F[x]_0^6$

$W = 14.1 \times 6$

$W =84.6\ Nm$

hence, work done by the force is equal to $W =84.6\ Nm$

7 0

3 years ago

A device for training astronauts and jet fighter pilots is designed to rotate the trainee in a horizontal circle of radius 11.0

kvv77 [185]

The velocity of the trainee is 29 m/s or 0.42 rev/s

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration (m / s²)v = final velocity (m / s)</em>

<em>u = initial velocity (m / s)</em>

<em>t = time taken (s)</em>

<em>d = distance (m)</em>

Centripetal Acceleration of circular motion could be calculated using following formula:

$\large {\boxed {a_s = v^2 / R} }$

<em>a = centripetal acceleration ( m/s² )</em>

<em>v = velocity ( m/s )</em>

<em>R = radius of circle ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Radius of horizontal circle = R = 11.0 m

Force Felt by the Trainee = F = 7.80w

<u>Unknown:</u>

Velocity of Rotation = v = ?

<u>Solution:</u>

$F = ma$

$F = m\frac{v^2}{R}$

$7.80w = m\frac{v^2}{R}$

$7.80mg = m\frac{v^2}{R}$

$7.80g = \frac{v^2}{R}$

$7.80 \times 9.8 = \frac{v^2}{11.0}$

$v^2 = 840.84$

$v \approx 29 ~m/s$

$\omega = \frac{v}{R}$ → in rad/s

$\omega = \frac{v}{2 \pi R}$ → in rev/s

$\omega = \frac{29}{2 \pi \times 11.0}$

$\omega \approx 0.42 ~ rev/s$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Uniform Circular Motion : brainly.com/question/2562955
Trajectory Motion : brainly.com/question/8656387

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

6 0

3 years ago

Read 2 more answers