All categories

3 years ago

A stunt driver drives a car horizontally off the edge of a cliff at 3.8m/s and reaches the water below 2.5s later.

Physics

1 answer:

andreyandreev [35.5K]3 years ago

4 0

A. The cliff was 30.7 m high
B. I also got 9.5 as the horizontal distance

Here is my work, I find making charts like this one to find knowns and unknowns can be helpful

You might be interested in

A block of mass M is attached to a spring of negligible mass and can slide on a horizontal surface along the x-direction, as sho

Hoochie [10]

Answer:

Because of the frictional force, the net force will oppose direction of the block and be directed towards the left even tho the spring exerts no force at this point

6 0

2 years ago

Which of the following statements best describes the second law of

Serga [27]

Answer:

Explanation:

Heat flows from hot to cold to lower the temperature of hot areas and increase temperature of cold areas. The end result is that the 2 areas have the same temperature, thus increasing entropy.

7 0

3 years ago

A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holdi

Elis [28]

Answer: $115.52\ N$

Explanation:

Given

Length of plank is 1.6 m

Force $F_1$ is applied on the left side of plank

Force $F_2$ is applied 43 cm from the left end O.

Mass of the plank is $m=13.7\ kg$

for equilibrium

Net torque must be zero. Taking torque about left side of the plank

$\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N$

Net vertical force must be zero on the plank

$\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N$

8 0

2 years ago

1. How much force is needed to accelerate a 66 kg skier<br> at 2 m/sec2?

Tju [1.3M]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 66 × 2

We have the final answer as

Hope this helps you

3 0

3 years ago

A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o

gulaghasi [49]

A. 0.77 A

Using the relationship:

$P=\frac{V^2}{R}$

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

$P_1=I^2 R_1$

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

$P_2=I^2 R_2$

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

$P=\frac{E}{t}$

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0

2 years ago