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3 years ago

Which statement is correct about the car?

Physics

2 answers:

JulijaS [17]3 years ago

6 0

Answer:

Explanation:

OOf we are doing this stuff atm

So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B

olga55 [171]3 years ago

5 0

Answer:

The car is moving forward

Explanation:

=> Forward Force = Force on the engine = 2000 N

=> Backward Force = Air Resistance + Friction = 800+500 = 1300 N

Since the forward force is more, the car moves in the forward directing fighting with the air resistance as well as friction.

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A star is born when gas and dust from a nebula become so dense and hot that nuclear fusion starts. Which of the following forces

Kitty [74]

Your Question:
<span>A star is born when gas and dust from a nebula become so dense and hot that nuclear fusion starts. Which of the following forces is responsible for the formation of a star?
A.
friction
B.
gravitation
C.
magnetism
D.
electromagnetic energy</span>
Our Teams Answer: B. Gravitation

Our Teams aim's to please and We are happy to help! Its the answer is wrong please notify us quickly for we can fix it quickly! Thanks :)

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8 0

2 years ago

Read 2 more answers

Complex carbohydrates, such as _____ and _____, provide the body with long-lasting energy.

Sauron [17]

complex carbohydrates, such as starches and fiber, provide the body with long-lasting energy.

Hope this helps!

7 0

2 years ago

Read 2 more answers

The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of

motikmotik

Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

$F\propto \dfrac{mgv^2}{r}$

$F=\dfrac{kmgv^2}{r}$

mg is the weight of the car

r is the radius of the curve

v is the speed of the car

Case 1.

F = 640 pounds

Weight of the car, W = mg = 2600 pound

Radius of the curve, r = 650 ft

Speed of the car, v = 40 mph

$640=\dfrac{k(2600)(40)^2}{650}$

k = 0.1

Case 2.

Radius of the curve, r = 750 ft

Speed of the car, v = 30 mph

$F=\dfrac{0.1\times 2600\times (30)^2}{750}$

F = 312 N

Hence, this is the required solution.

6 0

2 years ago

Which of the following unit is the unit of resistance? * 1 point A. A B. Ω C. V D. R

tekilochka [14]

Answer:

its b

Explanation:

ohm is the unit of resistance

5 0

2 years ago

For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at

Minchanka [31]

The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression.
In general, the potential energy of gravitational field is defined as:
$U=-G \frac{mM}{r}$
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
$F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}$
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
$g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg$
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
$U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}$
When we plug in the numbers we get:
$U=-4.99\cdot 10^{10} J$
The potential energy has to be equal to the kinetic energy.
$E_k=4.99\cdot 10^{10} J$

3 0

2 years ago