Which is a molecule? Select one: a. graphene b. NaCl c. Ne d. trail mix

The unit of force, Newton is a derived unit. (give reasons)

vitfil [10]

Answer:

$\boxed{ \sf{see \: below}}$

Explanation

The unit of force, Newton is a derived unit because it depends on three fundamental units. i.e kilogram , meter and second.

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6 0

3 years ago

Test your prediction through calculation for the situations of the clay bob and the bouncy ball. Assume each has a mass of 100 g

melamori03 [73]

Answer:

a) Δp = -2.0 kgm / s, b) Δp = -4 kg m / s

Explanation:

In this exercise the change in moment of a ball is asked in two different cases

a) clay ball, in this case the ball sticks to the door and we have an inelastic collision where the final velocity of the ball is zero

Δp = p_f - p₀

Δp = 0 - m v₀

Δp = - 0.100 20

Δp = -2.0 kgm / s

b) in this case we have a bouncing ball, this is an elastic collision, as the gate is fixed it can be considered an object of infinite mass, therefore the final speed of the ball has the same modulus of the initial velocity, but address would count

v_f = - v₀

Δp = p_f -p₀

Δp = m v_f - m v₀

Δp = m (v_f -v₀)

Δp = 0.100 (-20 - 20)

Δp = -4 kg m / s

6 0

3 years ago

g A high-energy photon turns into and electron and a positron. (A positron has exactly the same mass as the electron, but opposi

yawa3891 [41]

Answer:

2 m = E / c^2 where m is mass of electron

E = h v where v is the frequency ( nu) of the incident photon

E = h c / y where y is the incident wavelength (lambda)

2 m = h / (c y)

y = h / (2 m c) wavelength required

y = 6.62 * 10E-34 / (2 * 9.1 * 10E-31 * 3 * 10E8) m

y = 3.31 / 27.3 E-11 m

y = 1.21 E -12 m = .0121 Angstrom units

5 0

3 years ago

Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

6 0

3 years ago

A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on

g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J

6 0

3 years ago