What is 1.293 metric tons in grams

A stone was dropped off a cliff and hit the ground with a speed of 88 ft/s . What is the height of the cliff? (Use 32 ft/s 2 for

bulgar [2K]

To solve this problem we will apply the linear motion kinematic equations, which describe the change in velocity, depending on the acceleration and the distance traveled, that is,

$v_f^2 = v_i^2 +2ah$

Where,

$v_f$ = Final Velocity

$v_i$ = Initial Velocity

a = Acceleration

h = height

Our values are given as,

$v_f = 88 ft/s\\v_i = 0 ft /s\\a = 32 ft/s^2\\$

Replacing we have,

$vf^2 = vi^2 + 2*a*h$

$88^2 = 0 + 2*32*h$

$h= 121 ft$

Therefore the height of the cliff is 121ft

5 0

3 years ago

Which have different numbers of electrons?

nexus9112 [7]

Answer:

isotopes

Explanation:

6 0

3 years ago

Each of the four expansion models (recollapsing, critical, coasting, and accelerating) predict different ages for the universe,

cupoosta [38]

Answer:

This is because the age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.

Explanation:

The age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.

This is due to the fact that the expansion rate in the coasting model is constant and never changes. Because the cosmos is growing faster now than during the old days, recollapsing and critical models give shorter ages. According to the accelerating model, the universe is growing at a slower rate currently than in the past, implying an older age.

4 0

2 years ago

Martha must carry a 45 N package up three flights of stairs. Each flight of stairs has a height of 2.3m, and the actual distance

Darina [25.2K]

Answer:

310.5 J

Explanation:

The total work done by Martha is equal to the increase in gravitational potential energy of the package, which is equal to

$\Delta U = mg\Delta h$

where

(mg) = 45 N is the weight of the package

$\Delta h$ is the increase in height of the package

The package is carried up 3 flights of stairs, each one with a height of 2.3 m, so the total increase in heigth is

$\Delta h = 3 \cdot 2.3 m=6.9 m$

And so, the work done by Martha is

$U=(45 N)(6.9 m)=310.5 J$

3 0

3 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

2 years ago