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3 years ago

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Two small space probes have been slowed to 10m/s as they approach the moon from the same direction. Probe 1 has a mass of 86kg a

nd probe 2 has a mass of 100kg what is their total momentum

1 answer:

bulgar [2K]3 years ago

4 0

Answer:

B

Explanation:

B

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Now in "real life," this automobile is cruising at 20.5 m/s (equal to 73.8 km/hr) when it is about to hit a pedestrian stuck in

algol13

Answer:

He needs 1.53 seconds to stop the car.

Explanation:

Let the mass of the car is 1500 kg

Speed of the car, v = 20.5 m/s

He will not push the car with a force greater than, $F=2\times 10^4\ N$

The impulse delivered to the object is given by the change in momentum as :

$F\times t=mv\\\\t=\dfrac{mv}{F}\\\\t=\dfrac{1500\times 20.5}{2\times 10^4}\\\\t=1.53\ s$

So, he needs 1.53 seconds to stop the car. Hence, this is the required solution.

5 0

3 years ago

10. A change in

pogonyaev

Answer:

d velocity will be the one according to me

6 0

3 years ago

Describe how humans’ actions and natural processes have modified coastal regions in Louisiana and other locations

il63 [147K]

Coastal erosion has depleted a large portion of South Louisiana's wetlands along the coastline in swamps and marshes mainly due to storm surges. But other factors also contributed to this erosion. Canals and waterways dug through the marshes and swamps for the oil industry is one factor. Man-made levees erected to provide protection to residents living adjacent to the river is another major cause. Large scale logging especially in the early 1900's also damaged the wetlands.

3 0

4 years ago

A spring has a relaxed length of 35 cm (0.35 m) and its spring stiffness is 12 N/m. You glue a 66 gram block (0.066 kg) to the t

Sonbull [250]

We have that the value for y is

$y=0.12376m$

From the Question we are told that

Relaxed length of 35 cm (0.35 m)

Spring stiffness is 12 N/m.

Glue a 66 gram block (0.066 kg)

Total length is l_t=16 cm

Block during a 0.24-second interval

Generally the equation for the Force of spring of block is mathematically given as

$F_{spring}=kx\\\\\ F_{spring}=12*(0.35-0.16)\\\\ F_{spring}=2.28N$

Generally

$F_{elastic}=-mg\\\\F_{elastic}=-0.066*9.8\\\\F_{elastic}=-0.6468N$

Generally,Net Force

$F_{net}=F_{spring}-F_{elastic}\\\\F_{net}=2.28N-(-0.6468N)\\\\F_{net}=2.9268N$

Generally,Velocity of block

$V_y=\frac{py}{m}$

Where

$Py= F_{net}*dt\\\\Py= 2.9268*0.08\\\\Py=0.234144$

Therefore

$V_y=\frac{0.234144}{0.066}\\\\V_y=3.547m/s$

Generally the equation for the Velocity of block is mathematically given as

$Vy=\frac{yf-yt}{t-t_0}\\\\y=(3.547(0.08-0))-0.16$

$y=0.12376m$

For more information on this visit

brainly.com/question/22271063?referrer=searchResults

4 0

3 years ago

A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA. What amount of vars must be added to bring the pf to 0.85

kvasek [131]

Answer:

$\mathtt{Q_{sh} = 600.75 \ vars}$

Explanation:

Given that:

A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA

Here:

the initial power factor i.e cos θ₁ = 0.7 lag

θ₁ = cos⁻¹ (0.7)

θ₁ = 45.573°

Active power P = 1500 watts

Apparent power S = 2100 VA

What amount of vars must be added to bring the pf to 0.85

i.e the required power factor here is cos θ₂ = 0.85 lag

θ₂ = cos⁻¹ (0.85)

θ₂ = 31.788°

However; the initial reactive power $Q_1$ = P×tanθ₁

the initial reactive power $Q_1$ = 1500 × tan(45.573)

the initial reactive power $Q_1$ = 1500 × 1.0202

the initial reactive power $Q_1$ = 1530.3 vars

The amount of vars that must therefore be added to bring the pf to 0.85

can be calculated as:

$Q_{sh} = P( tan \theta_1 - tan \theta_2)$

$Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)$

$Q_{sh} = 1500( 1.0202 - 0.6197)$

$Q_{sh} = 1500( 0.4005)$

$\mathtt{Q_{sh} = 600.75 \ vars}$

3 0

3 years ago