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3 years ago

Two cars are moving with velocities 70km/hr and 50km/hr in east and west direction respectively.

Physics

1 answer:

viva [34]3 years ago

5 0

Answer:

70 + 50 = 120 km/hr

Explanation:

The driver of either car would see the other car approaching or departing at 120 kph

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You walk to the north, then turn 90° to your left and walk another How far are you from where you originally started?

rodikova [14]

Whatever distance north and then west you walked, you are then

(1.41 x that distance)

northwest of where you started.

3 0

3 years ago

A 20.0 kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at 34.0° above horizontal. A constant horizont

ankoles [38]

Answer:

987 joules, 3.01s

Explanation:

(A)

from the attached diagram

net force, Fnet, pulling the crate up the ramp is given by

Fnet = FcosФ - WsinФ - Fr

where FcosФ is the component of horizontal force 290N resolved parallel to the plane

WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane

Fr = constant opposing frictional force

Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65

Fnet = 240.421 - 109.602 - 65

Fnet = 65.82N

Work done on the crate up the ramp, W, is given by

W = Fnet × d (distance up the plane)

W = 65.819 × 15

W = 987.285 joules

W = 987 joules (to 3 significant Figures)

(B)

to calculate the time of travel up the ramp

we use the equation of motion

$s = ut + \frac{1}{2}at^{2}$

where s = distance up the plane, 15m

u = Initial velocity of the crate, which is 0 for a body that is initially at rest

a = acceleration up the plane, given by

$a = \frac{Fnet}{m}$

where m = mass of the crate, 20 kg

now, $a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}$

from, $s = ut + \frac{1}{2}at^{2}$

$15 = 0*t + \frac{1}{2}* 3.291 * t^{2}$

15 = 0 + 1.645 $t^{2}$

15 = 1.645 $t^{2}$

$t = \sqrt{\frac{15}{1.645} }$

$t = 3.019$

t = 3.01s (to 3 sig fig)

7 0

3 years ago

Bob is pushing a box across the floor at a constant speed of 1.7 m/s, applying a horizontal force whose magnitude is 70 N. Alice

OlgaM077 [116]

Answer:

a) 70 N, b) b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) A constant speed means that magnitude of friction force is equal to the magnitude of the external force. The friction force is directly proportional to the normal force, which is equal to the weight of the box. Therefore, the magnitude of the force is 70 N.

b) Alice used initially a greater force to accelerate the box up to needed speed and later reduced the external force to keep speed constant. The right choice is option b.

3 0

3 years ago

nexus9112 [7]

Answer:

As the skydiver falls to the Earth, she experiences positive acceleration only due to gravity.

Explanation:

As the skydiver falls to the Earth, she experiences friction in the form of air resistance which tries to slow her down and is proportional to the her velocity. So it cannot have a positive acceleration as it acts in opposite direction to slow her down.

Inertia during skydiving is experienced when we open an parachute, the parachute slows down the speed of are descent hence changing our inertia of motion with a velocity.

Only the Earth's, gravitational field has an positive acceleration as it pulls us towards the Earth, hence increasing our velocity.

3 0

4 years ago

Three objects are attached to a massless rigid rod that has an axis of rotation as shown. Assuming all of the mass of each objec

Vitek1552 [10]

The aggregate of all the given moment of inertia's will be the moment of inertia of this system.
as, moment of inertia is given as
l = m * r^2
so, finding the moment of inertia of all the individual and adding them

I=2∗1^2+1∗2^2+.5∗2.5^2
=9.125

3 0

3 years ago