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WINSTONCH [101]
3 years ago
5

Please help me with this Science Worksheet

Physics
1 answer:
sdas [7]3 years ago
7 0
When the mother is a Father is a dominant homozygous and the Mother is heterozygous
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Ahmad is riding his bicycle. He finds that he can accelerate from rest at 0.44 m/s^2 for 5 s to reach a speed of 2.2 m/s. The to
snow_lady [41]

Answer:

1) The force Christian can exert on his bicycle before picking up the the cargo is 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo is 647.46 N

Therefore, Christian has to exert more force on his bike after picking up the cargo

Explanation:

The given parameters are;

The mass of Christian and his bicycle = 54 kg

The mass of the cargo = 12 kg

1) The force Christian can exert on his bicycle before picking up the the cargo = Mass of Christian and his bicycle × Acceleration due to gravity

∴ The force Christian can exert on his bicycle before picking up the the cargo = 54 kg × 9.81 m/s² = 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo = (54 + 12) kg × 9.81 m/s² = 647.46 N

Therefore, Christian has to exert more force on his bike after picking up the cargo.

7 0
3 years ago
WILL GIVE BRAINLIEST!!!
Annette [7]

Answer:

Explanation:

Divergent Boundary

7 0
3 years ago
Read 2 more answers
Define average velocity.Immersive Reader
vagabundo [1.1K]

Answer:

The slope of a graph of position vs time

4 0
3 years ago
Kepler's First Law states that the shape of planetary orbits is a/an ________________ ?
harkovskaia [24]

Answer:

elliptical orbit

Explanation:

There are three laws of planetary motion, which are called Kepler's law of planetary motion.

First Law : It states that all the planets revolve around the sun in an elliptical path and the sun is at one focus of that elliptical path.

4 0
3 years ago
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

6 0
3 years ago
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