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lyudmila [28]
3 years ago
14

Science help please!

Physics
1 answer:
OLga [1]3 years ago
3 0

Answer:

104 N

Explanation:

m = 1300 kg

a = 0.08m/s^2

F = 1300*0.08

F = 104 N

Newtons is the unit of force.

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A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an
anyanavicka [17]

Answer:

The value of tangential acceleration \alpha_{t} =  40 \frac{m}{s^{2} }

The value of radial acceleration \alpha_{r} = 80 \frac{m}{s^{2} }

Explanation:

Angular acceleration = 50 \frac{rad}{s^{2} }

Radius of the disk = 0.8 m

Angular velocity = 10 \frac{rad}{s}

We know that tangential acceleration is given by the formula \alpha_{t} = r \alpha

Where r =  radius of the disk

\alpha = angular acceleration

⇒ \alpha_{t} = 0.8 × 50

⇒ \alpha_{t} = 40 \frac{m}{s^{2} }

This is the value of tangential acceleration.

Radial acceleration is given by

\alpha_{r} = \frac{V^{2} }{r}

Where V = velocity of the disk = r \omega

⇒ V = 0.8 × 10

⇒ V = 8 \frac{m}{s}

Radial acceleration

\alpha_{r} = \frac{8^{2} }{0.8}

\alpha_{r} = 80 \frac{m}{s^{2} }

This is the value of radial acceleration.

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Approximately how much air is in a column 1-cm2 in cross section that extends from sea level to the top of the atmosphere?
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Approximately 101 N air is in a column 1-cm2 in cross-section that extends from sea level to the top of the atmosphere

The basic level for determining height and depth on Earth is the sea level. The ocean's surface tends to seek the same level since it is one continuous body of water. However, the sea level is never fully level due to winds, currents, river discharges, and changes in gravity and temperature.

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2 years ago
Fusion and fission are both processes that involve which fundamental force? to ask? A. gravitational B. weak nuclear C. electrom
marin [14]
Fusion & Fission are both processes that involve D. Strong Nuclear Fundamental forces
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5 0
4 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s =  2500/(185644.54)

s=0.0134


So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0
3 years ago
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