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natima [27]
3 years ago
7

A truck is traveling due north at 50 km/hr approaching a crossroad. On a perpendicular road a police car is traveling west towar

d the intersection at 45 km/hr. Both vehicles will reach the crossroad in exactly one hour. Find the vector currently representing the displacement of the truck with respect to the police car.
Physics
1 answer:
kirill [66]3 years ago
8 0

Answer:45\hat{i}+50\hat{j}

Explanation:

Truck velocity=50 km/hr

police car velocity =45 km/hr

Let they start at t=0 s

After one hour

truck displacement=50\times 1=50 km

In vector Form(r_1)=50\hat{j}

For Police car displacement=45\times 1=45 km

in vector form(r_2)=-45\hat{i}

r_{12}=50\hat{j}-(-45\hat{i})

r_{12}=45\hat{i}+50\hat{j}

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What is the mass of 3 m3 of a substance having density 1200 kg/m3​
adell [148]

Answer:

3600 kg

Explanation:

From the question,

Density = Mass/Volume

D = M/V.............................. Equation 1

Where D = Density of the substance, M = mass of the substance, V = Volume of the subtance.

Make M the subject of the equation

M = D×V ............................ Equation 2

Given: D = 1200 kg/m³, V = 3 m³.

Substitute these values into equation 2

M = 1200×3

M = 3600 kg.

Hence the mass of the substance is 3600 kg

4 0
2 years ago
A bicycle pump contains 20 cm3 of air at a pressure of 100 kPa. The air is then pumped in a tyre of volume 100 cm3. Calculate th
Natasha2012 [34]

Answer:

The pressure of the air in the tyre is 20 kPa

Explanation:

The parameters for the bicycle pump and tyre are;

The volume of air contained in the bicycle pump, V₁ = 20 cm³

The pressure of the air contained in the bicycle pump, P₁ = 100 kPa

The volume (available) of the tyre, where the air is pumped, V₂ = 100 cm³

Let P₂ represent the pressure in the tyre after the air is pumped

By Boyle's law, we have that at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure;

Mathematically, Boyle's law gives the following equation;

P₁ × V₁ = P₂ × V₂

∴ P₂ = (P₁ × V₁)/V₂

Substituting the known values gives;

P₂ = (100 kPa × 20 cm³)/(100 cm³)

∴ P₂ = 100 kPa × 1/5 = 20 kPa

P₂ = 20 kPa

The pressure of the air in the tyre = P₂ = 20 kPa.

7 0
2 years ago
A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s
ICE Princess25 [194]

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

5 0
3 years ago
Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
lutik1710 [3]

Answer:

speed of electrons = 3.25 × 10^{7} m/s

acceleration in term g is 3.9 × 10^{17} g.

radius of circular orbit is 2.76 × 10^{-4} m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = \sqrt{\frac{2qV}{m}}

v = \sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}

v = 3.25 × 10^{7} m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = \frac{Bqv}{m}  

a =  \frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}

a = 3.82 × 10^{18} m/s²

and acceleration in term g

a = \frac{3.82\times 10^{18}}{9.81}  

a = 3.9 × 10^{17} g

acceleration in term g is 3.9 × 10^{17} g.

and

electron moving in circular orbit has centripetal force

F = \frac{mv^2}{r}  

Bqv = \frac{mv^2}{r}  

r = \frac{mv}{Bq}  

r = \frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}  

r = 2.76 × 10^{-4} m

radius of circular orbit is 2.76 × 10^{-4} m

8 0
3 years ago
The electric field in a particular thundercloud is 3.8 x 105 N/C. What is the acceleration (in m/s2) of an electron in this fiel
Svetradugi [14.3K]

Answer:

Answer:

6.68 x 10^16 m/s^2

Explanation:

Electric field, E = 3.8 x 10^5 N/C

charge of electron, q = 1.6 x 10^-19 C

mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of the electron.

The force due to electric field on electron is

F = q E

where q be the charge of electron and E be the electric field

F = 1.6 x 10^-19 x 3.8 x 10^5

F = 6.08 x 10^-14 N

According to Newton's second law

Force  = mass x acceleration

6.08 x 10^-14 = 9.1 x 10^-31 x a

a = 6.68 x 10^16 m/s^2

Explanation:

8 0
3 years ago
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