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2 years ago

What is the resistance force when you walk up an inclined plane? Please help quick!

1 answer:

5 0

Friction
Friction also affects the movement of an object on a slope. Friction is a force that offers resistance to movement when one object is in contact with another. Imagine now that you were on the downside of the object and applying force to keep the object in the same place (not moving)

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If a car changes ita velocity from 32km/hr to 54km/hr in 8.0 seconds, what is its acceleration

Andrews [41]

Firstly, let's convert the velocities in km/hr to m/s
32*1000/3600=8.89m/s
54*1000/3600=15m/s
From the formula, acceleration=V-U/t
15-8.89/8=0.76m/s²
hope this helps.

7 0

2 years ago

Which statement about allele combinations is accurate?

Pepsi [2]

Offspring get two alleles for each trait – one from each parent.

6 0

2 years ago

A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of

Step2247 [10]

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

$\eta = 1-\frac{T_L}{T_H}$

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

$\eta = 1-\frac{T_L}{T_H}$

$\eta = 1-\frac{293}{713}$

$\eta = 0.589$

Therefore the maximum possible efficiency the car can have is 58.9%

4 0

2 years ago

An object on Earth weighs 150 N. What is its mass?

Alex777 [14]

Answer:

$\tt \: mass \: = \frac{weight}{acceleration \: due \: to \: gravity}$

$\longrightarrow \tt \: \frac{150}{10}$

$\longrightarrow \boxed{ \tt{15 \: kg}}$

Our final answer is 15 kg .

----------- HappY LearninG<3 ------------

3 0

1 year ago

A test charge is placed at a distance of 2.5 × 10-2 meters from a charge of 6.4 × 10-5 coulombs. What is the electric field at t

Novosadov [1.4K]

Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance.

E = kQ / d²

k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m

Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²

Electric field at the test charge is 921600000 N/C

8 0

2 years ago

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