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Sophie [7]
3 years ago
9

A toy cart at the end of a string 0.70 m long moves in a circle on a table. The cart has a mass of 2.0 kg and the string has a b

reaking strength of 40.N. Calculate the maximum speed the cart can attain without breaking the string
Physics
1 answer:
luda_lava [24]3 years ago
8 0
Given that the mass of the toy cart is 2.0 kg and and the acceleration is unknown, the normal formula would be a=f/m where a is acceleration, f is force and m is mass but the string's breaking strength is 40n so I think the formula in this case will be f is greater than m*a
40 is greater than 2a
40 is greater than 2a
40/2 is greater than 2a/2
20m/s² is greater than a 
Therefore the maximum speed the toy cart should have should be less than 20m/s²
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Answer:

Explanation:

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5 0
3 years ago
Compare the gravitational force on a 33 kg mass at the surface of the Earth (with ra-
poizon [28]
The formula we will be using is;F = G m1 m2 / r^2 
F earth = 308N 
F moon = G me m2 / (81.3 * 0.27^2 RE^2) = 1/5.927 G me m2 / RE^2 = F earth / 5.927 = 52 N is the force of the earth
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4 0
3 years ago
An astronaut in a space craft looks out the window and sees an asteroid move pas a backward direction at 68 mph relative to the
Annette [7]

Answer: -194 mph

Explanation:

Taking into account the <u>Sun as the center </u>(origin, point zero) <u>of the reference system</u>, the velocity of the spacecraft relative to the Sun V_{R-S} is:

V_{R-S}=126mph  Note it is <u>positive</u> because the spacecraft is moving <u>away</u> from the Sun

Taking into account the <u>spacecraft as the center of another reference system</u>, the velocity of the asteroid relative to the spacecraft V_{A-R} is:

V_{A-R}=-68mph Note it is <u>negative</u> because the asteroid is moving<u> towards</u> the spacracft.

Now, the velocity of the asteroid relative to the Sun V_{A-S} is:

V_{A-S}=V_{A-R}-V_{R-S}

V_{A-S}=-68mph-126mph

Finally:

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6 0
3 years ago
Define Newton's third law:
Kisachek [45]

Answer:

the answer is B

Explanation:

the law states that for every action there is an equal and opposite reaction. action and reaction forces always act on two two different bodies and therefore THEY ALWAYS EXIST IN PAIRS.

3 0
3 years ago
A hot iron horseshoe (mass = 0.35 kg ), just forged, is dropped into 1.40 L of water in a 0.45 kg iron pot initially at 22.0°C.
olga_2 [115]

Answer:

420 °C

Explanation:

m_{h} = mass of the horseshoe = 0.35 kg

m_{w} = mass of the water = 1.40 L = 1.40 kg

m_{i} = mass of the iron pot = 0.45 kg

c_{i} = specific heat of iron = 450 J kg⁻¹ °C⁻¹

c_{w} = specific heat of water = 4186 J kg⁻¹ °C⁻¹

T_{hi} = initial temperature of the horseshoe = ?

T_{wi} = initial temperature of the water = 22 °C

T_{pi} = initial temperature of the iron pot = 22 °C

T_{f} = final temperature = 32 °C

Using conservation of Heat

m_{h}c_{i}(T_{hi} - T_{f}) = m_{w}c_{w}(T_{f} - T_{wi}) + m_{i}c_{i}(T_{f} - T_{pi})

(0.35)(450)(T_{hi} - 32) = (1.40)(4186)(32 - 22) + (0.45)(450)(32 - 22)

(157.5)(T_{hi} - 32) = 60629

T_{hi} - 32 = 384.95

T_{hi} = 420 °C

7 0
3 years ago
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