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3 years ago

WILL UPVOTE EVERY ANSWER! MULTIPLE CHOICE!

Physics

1 answer:

Ad libitum [116K]3 years ago

8 0

D = m /v
d = 30 / 0.5 Kg/m³
d = 60 Kg / m³

In short, Your Answer would be Option A

Hope this helps!

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You're driving down the highway late one night at 21 m/s when a deer steps onto the road 35 m in front of you. Your reaction tim

andreev551 [17]

Speed with which initially car is moving is 21 m/s

Reaction time = 0.50 s

distance traveled in the reaction time d = v t

d = 21 * 0.50 = 10.5 m

deceleration after this time = -10 m/s^2

now the distance traveled by the car after applying bakes

$v_f^2 - v_i^2 = 2a d$

$0 - 21^2 = 2(-10)d$

$d = 22.05 m$

so total distance moved before it stop

d = 22.05 + 10.5 = 32.55 m

so the distance from deer is 35 - 32.55 = 2.45 m

now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop

so our distance after brakes are applied is d = 35 - 10.5 = 24.5 m

again by kinematics

$v_f^2 - v_i^2 = 2 ad$

$0 - v^2 = 2(-10)(24.5)$

$v = 22.1 m/s$

so maximum speed would be 22.1 m/s

5 0

2 years ago

a car traveling at 30m/s notices an accident about 65m up the road. The car's brakes are capable of decelerating at a rate of 6.

uysha [10]

Yes because you would have at least 3 car spaces

4 0

2 years ago

Brainlist nd 20 point

Alborosie

the same is the answer

6 0

1 year ago

Write the realtionship between the density of a liquid and its upthrust? clarify

kipiarov [429]

Answer:

B = ρ g V_liquid

the thrust is proportional to the density of the liquid

Explanation:

The density of a liquid is defined as the relationship between the mass and the volume of the liquid

ρ = m / V

The upward push of the liquid is given by the principle of Archimedes Archimedes establishes that the push is equal to the weight of the dislodged liquid

B = W_liquid

B = m _liquid g

we substitute mass for density

B = ρ g V_liquid

therefore we see that the thrust is proportional to the density of the liquid

8 0

2 years ago

What are the horizontal and vertical velocities of a stunt bike that leaves a ramp at 100 km/hr and at an angle of 35 degrees?

Alex787 [66]

Horizontal velocity: 81.9 km/h

Vertical velocity: 57.4 km/h

Explanation:

We can solve this problem by resolving the velocity vector into its component along the horizontal and vertical direction.

The horizontal velocity of the stunt bike is given by:

$v_x = v cos \theta$

where

v = 100 km/h is the magnitude of the velocity

$\theta=35^{\circ}$ is the angle of projection

Substituting, we find

$v_x = (100)(cos 35^{\circ})=81.9 km/h$

The vertical velocity instead is given by

$v_y = v sin \theta$

where

$v=100 km/h$

$\theta=35^{\circ}$

Substituting,

$v_y = (100)(sin 35^{\circ})=57.4 km/h$

Learn more about vector components:

brainly.com/question/2678571

#LearnwithBrainly

4 0

2 years ago