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makkiz [27]
3 years ago
12

PLZZZZ HELP QUICK . 30 points. Emma weighs 560 N, she has decided to stay in shape so she is working out every morning, This mor

ning she ran and lifted weight! 1. Calculate Emma's mass ( Gravity g= 9.8 N/kg). 2. If she ran 350 meters in 65 seconds, what was her velocity? 3. When she lifted weights, she exerted a 35 N to lift one of the weight 2 meters over the ground, what is the amount of work she performed? Note that units are important, so do not give any number without a unit to avoid losing points.​
Physics
1 answer:
Phantasy [73]3 years ago
3 0

Force equals mass time acceleration. Weight is a force and it can replace force in the equation. The acceleration would be gravity, which is an acceleration.

1.)

Fw (weight) = m (mass) · g (gravity, 9.8 m/s²)

Fw = m * 9.81 m/s²

560N = m · 9.81 m/s²

m ≈ 57.08 kg

2.)

d = 350 meters

t = 65 seconds

velocity = d/t

velocity = 350 meters / 65 seconds

velocity ≈ 5.38 meters/sec

3.)

Force = 35N

Distance = 2 meters

Work = Force · Distance

Work = 35N · 2 meters

Work = 70 J

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3 years ago
A platinum ball weighing 100 g is removed from a furnace and dropped into 400 g of water at 0 degree C. If the equilibrium tempe
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Answer:

T = 1010 degree Celsius

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mass of water (Mw) = 400 g

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we can calculate the temperature of the furnace from the equation before

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4 years ago
Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo
djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

f_s = \mu_s N --- (1)

where;

f_s = static friction force

\mu_s = coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

N = ma_{min}

From equation (1)

f_s = \mu_s ma_{min}

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

F = f_s

mg = \mu_s ma_{min}

a_{min} = \dfrac{mg }{ \mu_s m}

a_{min} = \dfrac{g }{ \mu_s }

where;

\mu_s = 0.383 and g = 9.8 m/s²

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3 years ago
How small are the wavelengths of gamma ray radiation?
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3 years ago
Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57
Harlamova29_29 [7]
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
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In the second figure provided details are given on which are the angles in question, \Delta x is the distance between both rays.
\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705
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At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
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