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2 years ago

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For the instant represented, car has a speed of 100 km/h, which is increasing at the rate of 8 km/h each second. Simultaneously,

car B also has a speed of 100 km/h as it rounds the turn and is slowing down at the rate of 8 km/h each second. Determine the acceleration that car appears to have to an observer in car A.

1 answer:

GuDViN [60]2 years ago

4 0

Answer:

acceleration of car B appears at slowing down by 16 km/h each second

Explanation:

As we know by the concept of relative acceleration

$a_{BA} = a_B - a_A$

now we know that

Car A is moving with speed 100 km/h such that its speed is increasing at rate of 8 km/h each second

so acceleration of car A is given as

$a_A = 8 km/h/s$

now car B is also moving at same speed of 100 km/h but its speed is slowing down at rate of 8 km/h each second

so acceleration of car B is negative as it is decelerating

$a_B = - 8 km/h/s$

So here we have from above formula

$a_{BA} = a_B - a_A = - 8 - 8 = -16 km/h/s$

so acceleration of car B appears at slowing down by 16 km/h each second

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Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with the message 0.44 kg is on the table. Becc

jarptica [38.1K]

Calculate the weight of the table through the equation,

W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity. Substituting the known values,
W = (0.44 kg)(9.8 m/s²)
<em>W = 4.312 N</em>

The components of this weight can be calculated through the equation,

Wx = W(sin θ)

and Wy = W(cos θ)

x - component:
Wx = W(sin θ)
Substituting,
Wx = (4.312 N)(sin 150°) = <em>2.156 N</em>

Wy = (4.312 N)(cos 150°) =<em> -3.734 N</em>

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2 years ago

A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu

Vesnalui [34]

Answer:

The magnitude of the resultant acceleration is 2.2 $m/s^2$

Explanation:

Mass (m) of the sailboat = 2000 kg

Force acting on the sailboat due to ocean tide is $F_1$ = 3000N

Eastwards means takes place along the positive x direction

Then $F_{1x}$ = 3000N and $F_{1y}$ = 0

Wind Force acting on the Sailboat is $F_2$ = 6000N directed towards the northwest that means at an angle 45 degree above the negative x axis

Then

$F_{2x}$ = -(6000N) cos 45 degree = -4242.6 N

$F_{2y}$ = (6000N) cos 45 degree = 4242.6 N

Hence , the net force acting on the sailboat in x direction is

$F_x = F_{1x}+ F_{2x}$

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is

$F_y = F_{1y}+ F_{2y}$

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

$\sqrt{(1242.6)^2 + (4242.6)^2$

$\sqrt{(1544054.76) + (17999654.8)}$

$\sqrt{(19543709.5)^2}$

4420.8 N

F = ma

$a = \frac{F}{m}$

$a =\frac{4420.8}{ 2000}$

=2.2 $m/s^2$

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2 years ago

A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of

scZoUnD [109]

Refer to the diagram shown below.

Neglect wind resistance, and use g = 9.8 m/s².

The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

The pole vaulter comes to res after the pad compresses by 50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
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Answer: - 82 m/s² (or a deceleration of 82 m/s²)

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2 years ago

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Otrada [13]

Answer:

B. Solar energy

Explanation:

The water cycle is driven primarily by the energy from the sun. This solar energy drives the cycle by evaporating water from the oceans, lakes, rivers, and even the soil. Other water moves from plants to the atmosphere through the process of transpiration.

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2 years ago

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The moon’s orbital speed around Earth is 3.680 × 10^3 km/h. Suppose the moon suffers a perfectly elastic collision with a comet

Naily [24]

Answer:

Speed of comet before collision is

$v_{2_{i}}=-2.5\times10^{3}\quad km/h$

Explanation:

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Solution:

$mass \quad of\quad moon = m_{1}\\\\mass\quad of \quad comet = m_{2} = 0.5 m_{1}\\\\speed\quad of\quad moon\quad before\quad collision = v_{1_{i}}=3.680\times 10^3 km/h\\\\speed \quad of\quad moon\quad after\quad collision=v_{1_{f}} = -4.40 \times 10^2 km/h\\\\speed\quad of\quad comet\quad after\quad collision =v_{2_{f}} =5.740 \times 10^3 km/h$

Case is considered as partially inelastic collision, by conservation of momentum

$m_{1}v_{1_{i}}+m_{2}v_{2_{i}}=m_{1}v_{1_{f}}+m_{2}v_{2_{f}}\\\\m_{1}v_{1_{i}}+0.5m_{1}v_{2_{i}}=m_{1}v_{1_{f}}+0.5m_{1}v_{2_{f}}\\\\v_{1_{i}}+0.5v_{2_{i}}=v_{1_{f}}+0.5v_{2_{f}}\\\\v_{2_{i}}=2(v_{1_{f}}+0.5v_{2_{f}}-v_{1_{i}})\\\\v_{2_{i}}=2(-4.40 \times 10^2+0.5(5.740 \times 10^3)-3.680 \times 10^3 )\\\\v_{2_{i}}=-2.5\times10^{3}\quad km/h$

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2 years ago