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3 years ago

A car travels for 5.5 h at an average speed of 75 km/h. How far did it travel? T TE

Physics

1 answer:

ANEK [815]3 years ago

5 0

Answer:

$\boxed {\boxed {\sf 412.5 \ kilometers }}$

Explanation:

Distance is the product of speed and time.

$d=s*t$

The speed of the car is 75 kilometers per hour. It traveled for 5.5 hours.

$s= 75 \ km/hr \\t= 5.5 \ hr$

Substitute the values into the formula.

$d= 75 \ km/hr * 5.5 \ hr$

Multiply. Note that the hours will cancel each other out.

$d= 75 km * 5.5 \\d= 412.5 \ km$

The car travelled <u>412.5 kilometers.</u>

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satela [25.4K]

Answer:

Explanation:

I'm pretty sure I got it right

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3 years ago

An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,

Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity

$v=\sqrt{2.14^2+0.34^2}=2.17m/s$

Direction,

$\theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0$

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0

3 years ago

Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav

umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

$v_i = 4.0 \ m/s$

then,

$v_f=\frac{4.0}{2}$

$=2.0 \ m/s$

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ $a=\frac{v_f^2-v_i^2}{2s}$

On substituting the given values, we get

⇒ $=\frac{(2.0)^2-(4.0)^2}{2\times 18.2}$

⇒ $=\frac{4-8}{37}$

⇒ $=\frac{-4}{37}$

⇒ $=0.108 \ m/s^2$

As we know,

⇒ $f=ma$

and,

⇒ $\mu_rmg=ma$

⇒ $\mu_r=\frac{a}{g}$

On substituting the values, we get

⇒ $=\frac{0.108}{9.8}$

⇒ $=0.011$

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3 years ago

Which type of mirror would you use to cook a hot dog and why?

alexandr1967 [171]

As long as it’s a good mirror then any one of them is fine bc at the end of the day i’m getting a hot dog

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2 years ago

An ultrasound pulse used in medical imaging has a frequency of 5 MHz and a pulse width of 0.5 us. Ap- proximately how many osci

podryga [215]

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3 years ago