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2 years ago

8

What are 2applications of radar that you have heard of in your life? What can radar tell an observer about some object, whether

it is an airplane, a car, a cloud?

1 answer:

RoseWind [281]2 years ago

4 0

Answer:

Radar is used in space and in military

Explanation:

Radar helps to identify the type of forget that is in military it can be used to identify the enemies. It helps to identify the position of the object i.e helps to determine how far the object or subject is. It can be used to guide objects in space and monitor movements of other space objects.

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A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point

Nonamiya [84]

Answer:

1.52 m

Explanation:

We are given that

Maximum force=F=679 N

Mass of rock ,m=385 kg

Distance,d=0.233 m

We have to find the minimum total length L of the rod required to move the rock.

Torque on rock= $T_1=mgd=385\times 9.8\times 0.233=879.11 N$

Where $g=9.8m/s^2$

Torque on man= $T_2=Force\times distance=(L-d)\times 679=(L-0.233)\times 679$

$T_1=T_2$

$(L-0.233)\times 679=879.11$

$L-0.233=\frac{879.11}{679}=1.29$

$L=1.29+0.233=1.523\approx 1.52 m$

Hence, the minimum total length of rod=1.52 m

3 0

3 years ago

An investigation has been completed similar to the one on latent heat of fusion, where steam is bubbled through a container of w

allochka39001 [22]

Answer:

$Q_a=330 cal$

$Q_w=6000cal$

Explanation:

From the question we are told that

Mass of the aluminum container 50 g

Mass of the container and water 250 g

Mass of the water 200 g

Initial temperature of the container and water 20°C

Temperature of the steam 100°C

Final temperature of the container, water, and condensed steam 50°C

Mass of the container, water, and condensed steam 261 g

Mass of the steam 11 g Specific heat of aluminum 0.22 cal/g°C

a) Heat energy on container

Generally the formula for mathematically solving heat gain

$Q_c=M_c *C_c*( \triangle T)$

Therefore imputing variables we have

$Q_a=50g *0.22*50-20$

$Q_a=330 cal$

b) Heat energy on water

Generally the formula for mathematically solving heat gain

$Q_w=M_w *C_w*( \triangle T)$

Therefore imputing variables we have

$Q_w=200 *1* 50-204$

$Q_w=6000cal$

7 0

3 years ago

What is the equivalent resistance of the circuit?

Ivanshal [37]

Answer:

Since the wire is not splitting at any point in the circuit,

the resistors are in series

Hence, Equivalent resistance = 10 + 20 + 30

Equivalent Resistance = 60 Ω

8 0

3 years ago

Listening to the radio, you can hear two stations at once. Describe this wave interaction

Sedaia [141]

Actually says Pantazis, since their frequencies are so wildly different, brain waves don’t interfere with radio waves. Even if that was the case, brain waves are so weak, they are hardly measurable at all. For comparison, says Pantazis, “the magnetic field of the earth is just strong enough to move the needle of a compass. Signals from the brain are a billionth of that strength.”

8 0

3 years ago

Read 2 more answers

Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra

vredina [299]

Answer:

At low pressure- $\mu_{k}=0.02315$

At high pressure- $\mu_{k}=0.00445$

Explanation:

Initial speed, $V_{i}=3.3 m/s$

Final speed, $V_{f}=3.3/2= 1.65 m/s$

Net horizontal force due to rolling friction $F_{net}=\mu_{k}$ mg where m is mass, g is acceleration due to gravity, $\mu_{k}$ is coefficient of rolling friction

From kinematic relation, $V_{f}^{2}- V_{i}^{2}=2ad$

For each tire,

$V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd$

Making $\mu_{k}$ the subject

$\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}$

Under low pressure of 40 Psi, d=18 m

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315$

Therefore, $\mu_{k}=0.02315$

At a pressure of 105 Psi, d=93.7

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445$

Therefore, $\mu_{k}=0.00445$

4 0

3 years ago