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3 years ago

The Washington Monument is 169.20 meters tall.

Physics

1 answer:

bonufazy [111]3 years ago

8 0

Answer: K.E = 8.3 J

Explanation:

The given parameters are

Height h = 169.2 m

Mass m = 0.005kg

From conservation of energy, at the top of the monument, the total energy will be kinetic energy which is equal to the potential energy. That is,

K.E = P.E

And

P.E = mgh

Where g = 9.81 m/s^2

Substitutes all the parameters into the formula

P.E = 0.005 × 9.81 × 169.2

P.E = 8.2908 J

Therefore, the kinetic energy (KE) of the nickel in her shirt pocket at the top of the monument is approximately 8.3 Joule.

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Indicate the result of each of the following unit vector cross products (unit vector hat-symbols not shown): . kxi = . jxi= -jxk

notka56 [123]

Answer:

Explanation:

The cross product of two vectors is given by

$\overrightarrow{A}\times \overrightarrow{B}=\left | \overrightarrow{A} \right |\left | \overrightarrow{B} \right |Sin\theta \widehat{n}$

Where, θ be the angle between the two vectors and \widehat{n} be the unit vector along the direction of cross product of two vectors.

Here, K x i = - j

As K is the unit vector along Z axis, i is the unit vector along X axis and j be the unit vector along axis.

The direction of cross product of two vectors is given by the right hand palm rule.

So, k x i = j

j x i = - k

- j x k = - i

i x i = 0

4 0

3 years ago

The next four questions refer to the situation below.

Anna11 [10]

Answer:

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}, t_{out} = $\frac{D}{v_s +v_r}$

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

$v_{sg 1} = v_{sr} + v_{rg}$

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

$v_{sg1}$ = D / $t_{out}$

D = v_{sg1} t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

$v_{sg 2} = v_{sr} - v_{rg}$

$v_{sg 2}$ = D / $t_{in}$

D = v_{sg 2} t_{in}

with the distance is the same we can equalize

$v_{sg1} t_{out} = v_{sg2} t_{in}$

t_{out} = t_{in}

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}

This must be the answer since the return time is known. If you want to delete this time

t_{in}= D / $v_{sg2}$

we substitute

t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

t_{out} = $\frac{D}{v_s +v_r}$

7 0

2 years ago

Biker A is cruising at a speed of 10.0 m/s when she passes biker B who is at rest at the origin of the coordinate system. Biker

Burka [1]

Answer: attached

Explanation:

3 0

3 years ago

The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a

mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

3 0

3 years ago

Read 2 more answers

Two blocks of ice, one four times as heavy as the other, are at rest on a frozen lake. A person pushes each block the same dista

qaws [65]

Answer:b

Explanation:

Given

mass of heavy object is 4m

mass of lighter object is m

A person pushes each block with same force F

According to Work Energy theorem Change in kinetic energy of object is equal to Work done by all the object

As launching velocity is same for both the object so heavier mass must possess greater kinetic energy . For same force heavier mass must be pushed 4 times farther than the light block .

$\Delta (K.E.)_H=\frac{1}{2}(4m)v^2$

$\Delta (K.E.)_L=\frac{1}{2}(m)v^2$

$\Delta K.E.=F\times d$

So the correct option is b

4 0

3 years ago