All categories

3 years ago

Suppose a 1300 kg car is traveling around a circular curve in a road at a constant

Physics

1 answer:

Scrat [10]3 years ago

4 0

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :

$F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N$

So, the force has a magnitude of 4212 N

You might be interested in

PLEASE HELP!!!!

Zina [86]

D is the answer
Imagine the magnetic field, it emanates from both

4 0

4 years ago

Two ships leave a harbor at the same time, traveling on courses that have an angle of 110∘ between them. If the first ship trave

Allushta [10]

Answer:

49.07 miles

Explanation:

Angle between two ships = 110° = θ

First ship speed = 22 mph

Second ship speed = 34 mph

Distance covered by first ship after 1.2 hours = 22×1.2 = 26.4 miles = b

Distance covered by second ship after 1.2 hours = 34×1.2 = 40.8 miles = c

Here the angle between the two sides of a triangle is 110° so from the law of cosines we get

a² = b²+c²-2bc cosθ

⇒a² = 26.4²+40.8²-2×26.4×40.8 cos110

⇒a² = 2408.4

⇒a = 49.07 miles

6 0

3 years ago

Irregular galaxies are very large. True or False?

xenn [34]

It's false i hope this helps :)

5 0

3 years ago

Read 2 more answers

After all that you have learned in this unit, construct a pamphlet (brochure) in Microsoft Publisher helping new freshman to enc

aleksandr82 [10.1K]

You're not going to like this answer, but it's the only one possible:. It wasn't I who learned anything in this unit. If it was either of us, it was YOU. I can't even tell from reading the question what the topic of the unit was. Was it pamphlets ? Microsoft Publisher ? Freshmen ? Getting Through High School ? This is a lot like asking me to write something "in your own words".

5 0

3 years ago

A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa

DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s$

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

3 0

4 years ago