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3 years ago

#9 help please I don’t know what formula to use

Physics

1 answer:

butalik [34]3 years ago

4 0

Imagine if the tourist could just get in the car right before the bear could catch him. This means they took the same amount of time to reach the car.

distance travelled by the tourist = d m

distance travelled by the bear = (26 + d) m

time taken by the tourist = time taken by the bear

$\frac{d}{4} =\frac{26+d}{6} \\6d=4(26+d)\\2d=104\\d=52$

So, the max d is 52.

Hope this helps:)

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Ice-skater slides toward a sled sitting on the ice and hits it. The skater exerts a 12.6 N force on the sled at an angle of 15.3

RideAnS [48]

Answer:

Expression of work done is

$W = Fd cos\theta$

Work done to move the sled is given as 187.2 J

Explanation:

As we know that the formula of work done is given as

$W = Fd cos\theta$

here we know that

F = 12.6 N

d = 15.4 m

$\theta = 15.3 degree$

so we will have

$W = 12.6 \times 15.4 cos15.3$

$W = 187.2 J$

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3 years ago

A racecar drives at a constant speed down a straight track. The car is in _?_

vagabundo [1.1K]

Answer:

the answer is

the car is in motion

7 0

3 years ago

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

5 0

4 years ago

Gulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells.

kumpel [21]

Answer:

v = 17.71 m / s

Explanation:

We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity

v² = v₀² - 2 g (y -y₀)

v² = 0 - 2g (y -y₀)

when it hits the stone the height is zero and part of the height of the seagull I

v² = 2g y₀

v = Ra (2g i)

let's calculate

v =√ (2 9.8 16)

v = 17.71 m / s

8 0

4 years ago

ASAP

scoundrel [369]

Answer:

Explanation:

Hooke's law! F(spring)=-kx

There's no tricky square law here. The spring constant doesn't change, only x (distance stretched) changes. Therefore, if distance is halved, Force will be halved.

5 0

3 years ago