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3 years ago

Please help I can’t get an answer

Physics

2 answers:

Anna007 [38]3 years ago

5 0

I am not 100% sure, more like 90% sure, that the answer is A.

Hope this helps! :)

svp [43]3 years ago

3 0

C is the answer glad to help

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3. What is the mass of a paratrooper who experiences an air resistance of 400 N and an acceleration of 4.5 m/s2

goblinko [34]

Answer:

88.89kg

Explanation:

The formula for mass is m=F/a. If we plug in the values, we get m=400N/4.5m/s^2. The mass is 88.89kg. We know that the unit is in kg because one newton (N) is 1kg*m/s^2. The m/s^2 is cancelled out by the acceleration, and we are left with kg.

4 0

3 years ago

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.9 kg and length L = 4.88 m to a uniform sphere with m

Bumek [7]

Answer:

a) total moment of inertia is 1359.05 kg m^2

b) angular acceleratio is 0.854rad/sec^2

Explanation:

Given data:

m1=6.9 kg

L=4.88 m

m2=34.5 kg

R=1.22 m

we klnow that moment of inertia for rod is given as

J1=(1/12) ×m×L^2

$J1 = (1/12) \times 6.9 \times 4.88^2 = 13.693 kg m^2$

moment of inertia for sphere is given as

J1=(2/5) ×m×r^2

$J1 = (2/5) \times 34.5 \times 1.22^2 = 20.539 kg m^2$

As object rotates around free end of rod then for sphere the axis around what it rotates is at a distance of d2=L+R

For rod distance is d1=0.5*L

By Steiner theorem

for the rod we get $J_1'=J_1 + m_1\times d_1^2$

$J_1' = 13.693 + 2.44^2\times 6.9 = 54.77 kg m^2$

for the sphere we get $J_2' = J_2 + m_2\times d_2^2$

$J2' = 20.539 + 34.5\times 6.1^2 = 1304.28 kg^m2$

And the total moment of inertia for the first case is

$J_{t1} = J_1'+J_2' = 54.77 + 1304.28 = 1359.05kg.m^2$

b) F=476 N

The torque for system is given as

$M = F\times d\times sin(a)$

where a is angle between Force and distance d

and where d represent distance from rotating axis.

In this case a = 90 degree

$M = F\times L/2$

M=476*2.44 = 1161.44 Nm

The acceleration is calculated as

$a_1 = \frac{M}{J_{t1}}$

$= \frac{1161.44}{1359.05}$

= 0.854 rad/sec^2

4 0

3 years ago

A 20-kg block slides down a fixed rough curved track The block has a speed of 5 0 m/s after its height above a horizontal surfac

Crank

Answer:

U = 102.8 J (100 J to two significant digits)

Explanation:

potential energy converted = 20(9.8)(1.8) = 352.8 J

kinetic energy at base of track = ½(20)5.0² = 250 J

energy (work) of friction 352.8 - 250 = 102.8 J

8 0

3 years ago

Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message e

Grace [21]

Answer:

The duration is $T =72 \ years /tex]Explanation:From the question we are told that The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m$

Generally the time it would take for the message to get the the other civilization is mathematically represented as

$t = \frac{D}{c}$

Here c is the speed of light with the value $c = 3.08 *10^{8} \ m/s$

=> $t = \frac{3.311 *10^{17} }{3.08 *10^{8}}$

=> $t = 1.075 *10^9 \ s$

converting to years

$t = 1.075 *10^9 * 3.17 *10^{-8}$

$t = 34 \ years$

Now the total time taken is mathematically represented as

$T = 2* t + 2 + 2$

=> $T = 2* 34 + 2 + 2$

=> [tex]T =72 \ years /tex]

4 0

3 years ago

What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a

Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

$M=-\frac{f_o}{f_e}$

Here, M is the magnification of an astronomical telescope, $f_e$ is the focal length of the eyepiece lens and $f_o$ is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put $f_o=74 cm$ and $f_e=2.4 cm$ in the above expression.

$M=-\frac{74}{2.4}$

M=-30.83

Therefore, the magnification of an astronomical telescope is -30.83.

5 0

4 years ago