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alex41 [277]
3 years ago
12

A hungry lion is looking for food. It went 5 miles (mi) before it spotted a zebra to stalk. It took this lion 0.2 hours to get t

o the zebra. What is the lion’s speed?
Question 1 options:

1 mi


1 mi/h


25 mi


25 mi/h
Physics
1 answer:
Helen [10]3 years ago
5 0

To solve this exercise, the following speed equation must be applied:

 V = d / t (mi/hr)

 V: speed (mi/hr)

 d: distance (mi)

 t: time (s)

We know the distance (d= 5 miles) and the time (t= 0.2 hours), so we proceed to replace these values in the formula to calculate the lion’s speed:

 V = d / t

 V = 5 mi/0.2 hr

 V = 25 mi/hr

 The lion’s speed is 25 mi/hr, so the last option is the right answer

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Select. The star cycle that is accurate
melomori [17]

One stellar-mass star, red giants, white dwarfs, or planetary nebulae will make up the correct star cycle. Option C is correct.

<h3 /><h3>What is the solar system?</h3>

The satellites of the planet, countless comets, asteroids, and meteoroids, as well as the interplanetary medium, make up the solar system.

The complete question is;

"Choose the correct star life cycle.

A. Supernova, star as well as red giant, the nebula is incorrect.

B. Nebula, white dwarf, planetary nebula That is incorrect.

C. Planetary nebula, red giant, white dwarf, and star with a single stellar mass That is true!

D. Nebula, a star of one stellar mass, is not correct. It is correct if the star had four or more stellar masses."

Star of one stellar mass, red giant, white dwarf, and planetary nebula make up the proper life cycle.

Hence option C is correct.

To learn more about the solar system refer to the link;

brainly.com/question/1207587

#SPJ1

5 0
2 years ago
A 18-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 30 N. Starting from rest, the sle
k0ka [10]

Answer:

Coefficient of kinetic friction = 0.146

Explanation:

Given:

Mass of sled (m) = 18 kg

Horizontal force (F) = 30 N

FInal speed (v) = 2 m/s

Distance (s) = 8.5 m

Find:

Coefficient of kinetic friction.

Computation:

Initial speed (u) = 0 m/s

v² - u² = 2as

2(8.5)a = 2² - 0²

a = 0.2352 m/s²

Nweton's law of :

F (net) = ma

30N - μf = 18 (0.2352)

30 - 4.2336 = μ(mg)

25.7664 =  μ(18)(9.8)

μ = 0.146

Coefficient of kinetic friction = 0.146

6 0
4 years ago
A 290 gg bird flying along at 6.2 m/sm/s sees a 9.0 gg insect heading straight toward it with a speed of 34 m/sm/s (as measured
Murrr4er [49]

Answer:

The bird's speed immediately after swallowing is 4.98 m/s.

Explanation:

Given that,

Mass of bird = 290 g

Speed = 6.2 m/s

Mass of sees = 9.0 g

Speed = 34 m/s

We need to calculate the bird's speed immediately after swallowing

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3}

Put the value into the formula

0.290\times6.2+0.009\times(-34)=(0.290+0.009)\times v_{3}

v_{3}=\dfrac{0.290\times6.2-0.009\times34}{(0.290+0.009)}

v_{3}=4.98\ m/s

Hence, The bird's speed immediately after swallowing is 4.98 m/s.

6 0
3 years ago
Rank the angular speed of the following objects, from highest to lowest:- A bowling ball of radius 12.3 cm rotating at 8.21 radi
dlinn [17]

Answer:

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

Explanation:

<u>To rank the angular speed (ω) of the objects, we need first calculate its value for every object:</u>

A bowling ball of radius 12.3cm rotating at 8.21 radians per second:

ω = 8.21 rad/s

A tire of radius 0.321m rotating at 75.8 rpm:

\omega = 75.8 \frac{rev}{min}\cdot \frac{2\pi rad}{1rev}\cdot \frac{1min}{60s} = 7.94rad/s

A 6.84cm diameter top spinning at 375 degrees per second:

\omega = 375 \frac{^\circ}{s}\cdot \frac{2\pi rad}{360^ \circ} = 6.54rad/s    

A square with sides (b) 0.458m long, whose corners are moving with tangential speed (v) 2.51 m/s as it rotates about its center:

\omega = \frac{v}{r} = \frac{v}{\frac{b}{2}\sqrt 2} = \frac{2.51 m/s}{\frac{0.458 m}{2} \sqrt 2} = 7.75rad/s

A rock on a string, being swung in a circle of radius 0.521 m with a centripetal acceleration (a) of 25.4 m/s²:

\omega = \sqrt \frac{a}{r} = \sqrt \frac{25.4 m/s^{2}}{0.521m} = 6.98rad/s

<u>Now, the rank of the angular speed of the objects, from highest to lowest is: </u>

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s) 

I hope it helps you!      

4 0
3 years ago
1.5-m length of straight wire experiences a maximum force of 1.2 N when in a uniform magnetic field that is 1.8 T. 1) What curre
Iteru [2.4K]

Answer:

 I = 0.44 A

Explanation:

The magnetic force on a conductor is given by the expression

       F = I L x B

Where bold letters indicate vectors, I is the current, L is the vector in the direction of the current, and B is the magnetic field

Since the force is maximum, the wire must be perpendicular to the magnetic field, therefore

        F = I L B sin 90

        I = F / L B

Let's calculate

        I = 1.2 / 1.5 1.8

        I = 0.44 A

7 0
3 years ago
Read 2 more answers
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