Ask question

Ask question

All categories

2 years ago

15

A person takes a trip, driving with a constant speed of 99.5 km/h, except for a 26.0-min rest stop. The person's average speed i

s 73.6 km/h. a) How much time is spent on the trip? (in h) b) How far does the person travel? (in km)

1 answer:

Korvikt [17]2 years ago

7 0

Answer:

1.65 h

121.39 km

Explanation:

Given that

speed of the driver, v = 99.5 km/h

time spent resting, t = 26 min

Average speed of the driver = 73.6 km/h

check attachment for calculation and how I arrived at the answer

You might be interested in

What type of galaxy is M82 based on its appearance in the visible-light view?

SSSSS [86.1K]

<h2>Answer: irregular</h2>

According to Hubble galaxies are classified into elliptical, spiral and irregular.

It should be noted this classification is based only on the visual appearance of the galaxy, and does not take into account other aspects, such as the rate of star formation or the activity of the galactic nucleus.

The classification is as follows:

1. Elliptical galaxies: Their main characteristic is that the concentration of stars decreases from the nucleus, which is small and very bright, towards its edges. In addition, they contain a large population of old stars, usually little gas and dust, and some newly formed stars.

2. Spiral galaxies: They have the shape of flattened disks containing some old stars and also a large population of young stars, enough gas and dust, and molecular clouds that are the birthplace of the stars.

3. Irregular Galaxies: Galaxies that do not have well-defined structure and symmetry.

In this context, galaxy M82 does not match with the first two types of galaxies, because it has not a defined shape.

Therefore, M82 is an irregular galaxy.

4 0

2 years ago

The gravitational force between two objects has a magnitude of F. If both masses were doubled and the distance between them doub

Fofino [41]

Answer:

F' = F

Explanation:

The gravitational force of attraction between two objects can be given by Newton's Gravitational Law as follows:

$F = \frac{Gm_1m_2}{r^2}$

where,

F = Force of attraction

G = Universal gravitational costant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects

Now, if the masses and the distance between them is doubled:

$F' = \frac{G(2m_1)(2m_2)}{(2r)^2}\\\\F' = \frac{Gm_1m_2}{r^2}$

<u>F' = F</u>

7 0

2 years ago

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

So

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

3 0

2 years ago

A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-

Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

$v_{s_{x}}=0$

$v_{s_{y}}=2.0\ m/s$

The velocity of the boat in term x and y coordinate

For x component,

$v_{b_{x}}=v_{b}\cos\theta$

Put the value into the formula

$v_{b_{x}}=5.60\cos19$

$v_{b_{x}}=5.29\ m/s$

For y component,

$v_{b_{y}}=v_{b}\sin\theta$

Put the value into the formula

$v_{b_{y}}=5.60\sin19$

$v_{b_{y}}=1.82\ m/s$

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

$v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}$

Put the value into the formula

$v_{sb_{x}=0-5.29$

$v_{sb}_{x}=-5.29\ m/s$

For y component,

$v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}$

Put the value into the formula

$v_{sb_{x}=2.-1.82$

$v_{sb}_{x}=0.18\ m/s$

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

7 0

2 years ago

Two pure tones are sounded together. The drawing shows the pressure variations of the two sound waves, measured with respect to

amm1812

Answer:

4.13Hz

Explanation:

f1 = 1/t1 = 1/0.022 = 45.45 Hz

f2 = 1/t2 = 1/0.0242= 41.32 Hz

No. of beats

= 45.45- 41.32

~ 4.13Hz

7 0

2 years ago