1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
koban [17]
2 years ago
8

a car has a velocity of 10ms-1. it accelerates at 0.2 ms-2 for half minute find the total distance travelled and final velocity

of car.​
Physics
1 answer:
Readme [11.4K]2 years ago
5 0

Answer:

Final velocity = 16 m/s

Total distane travelled = 390 m

Explanation:

We can use equation of motion to solve this:

v = u + at \\ v = 10 + 0.2(30) \\ v = 16 {ms}^{ - 1}

s = ut +  \frac{1}{2} a {t}^{2}  \\ s = 10(30) +  \frac{1}{2} (0.2) {(30)}^{2}  \\ s = 390m

You might be interested in
You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b
Radda [10]

Answer:

\theta=34 \textdegree

Explanation:

From the question we are told that:

Mass m=55kg

Angle \theta =28.0

Coefficient of static friction \alpha =0.680

Generally, the equation for Newtons second Law is mathematically given by

For

\sum_y=0

N=mgcos \theta

for

\sum_x=0

F_{s}=mgsin\theta

Where

F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta

F_{s}=0.68*55*9.8*cos 28

F_{s}=323.62N

Therefore

\alpha mgcos \theta=mg sin \theta

\theta=tan^{-1}(0.68)

\theta=34 \textdegree

6 0
2 years ago
A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at
stellarik [79]

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft

Also, at any instant t

\Rightarrow l^2=x^2+y^2

differentiate w.r.t.

\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s

5 0
3 years ago
A 2kg object initially going 4m/s to the right is later going 8m/s. whats the change in velocity?
Nezavi [6.7K]

The change in velocity is +4 m/s to the right (or -4 m/s to the left).

The object's mass is irrelevant.

6 0
3 years ago
A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i
marshall27 [118]

Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

μ_k = 0.646

d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

K = 89.39 - 65.59 = 23.8J

3 0
2 years ago
A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
3 years ago
Other questions:
  • A singly charged positive ion that has a mass of 6.07 ? 10-27 kg moves clockwise with a speed of 1.18 ? 104 m/s. the positively
    7·1 answer
  • If B is added to C, the result is a vector in the direction of the positive y-axis with a magnitude equal to that of If C = zi +
    9·1 answer
  • How much heat, in joules and in calories, must be added to a 75.0–g iron block with a specific heat of 0.449 j/g °c to increas
    6·1 answer
  • Please please please please PLEASE help!!!
    14·1 answer
  • Two plane mirrors are facing each other. They are parallel, 6 cm apart, and 24 cm in length, as the drawing indicates. A laser b
    12·2 answers
  • An object has a kinetic energy of 14 J and a mass of 17 kg , how fast is the object moving?
    5·2 answers
  • Calculate the weight of a new fast-food sandwich that has a mass of 0.1 kg (approximately the mass of a quarter pound). Think of
    10·1 answer
  • An infinitely long line charge of uniform linear charge density λ = -3.00 µC/m lies parallel to the y axis at x = -3.00 m. A poi
    10·1 answer
  • Please help meh need answers
    15·1 answer
  • According to Dr. paul Narguizian professor of Biology and Science Education at California State University, ______ are generaliz
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!