All categories

3 years ago

A car travels 2.5 hours in a northernly direction for 300 km. determine the cars speed and velocity.

Physics

1 answer:

wolverine [178]3 years ago

5 0

Answer:

Speed= 120 km/h

Velocity= 120 Km/h North

Explanation:

Time= 2.5 hours

Distance= 300 Km

Direction= North

Speed= distance/speed

Velocity= speed and direction

Substituting 300 km for speed and 2.5 hours for time then speed= 300/2.5= 120 km/h

The velocity= 120 km/h North

You might be interested in

Paul is driving on a sunny day what glasses should he wear

Natalka [10]

Sunglasses to shield from the sun

6 0

3 years ago

Read 2 more answers

A wagon is pulled at a speed of 0.40metets/seconds by a horse exerting an 1,800 Newton's horizontal force. what is the power of

Tatiana [17]

Given:

speed of 0.40meters/seconds

1,800 Newton's horizontal force

Required:

Power of the horse

Solution:

P = F(D/T) where P is power in watts, F is the force, D is the distance and T is time

P = (1,800N) (0.40 meters/seconds)

P = 720 Watts

6 0

3 years ago

A distance of 2.00 mm separates two objects of equal mass. If the gravitational force between them is 0.0104 N, find the mass of

Reptile [31]

F = $\frac{gm,m2}{r2}$
r = 0.002

0.0104 = $\frac{(6.673 x 10^{-11} }{ 0.002)^{2} }$

m = 25.0kg

4 0

3 years ago

Read 2 more answers

Free moneybbbbbbbbbbbbbbbbbbbbb

Lunna [17]

Answer:

Thank you very much✨

3 0

3 years ago

Read 2 more answers

A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block $m = 6.60$ kg

Initial speed of block $v _{i} = 1.56$ $\frac{m}{s}$

Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

$\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}$

$d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}$

$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0

3 years ago