All categories

2 years ago

Plz answer all 5 of the questions i will give brainlyest

Physics

1 answer:

lubasha [3.4K]2 years ago

6 0

Answer:

1)force:- force is an external affert by any body on an object.

2) force and motion are related to each other. force helps to create motion in any object.

3) frictional force

4) Newton

You might be interested in

If the car went 30 km west in 25 min. and then 40 km east in 35 min., what would be its total displacement?

Mandarinka [93]

The total displacement is 10 km.

8 0

2 years ago

In a laboratory, you determine that the density of a certain solid is 5.23× 10 −6 kg/m m 3 . convert this density into kilograms

tamaranim1 [39]

Density is given as

$\rho = 5.23 * 10^{-6} kg/mm^3$

now we have to convert this density into $kg/m^3$

now we have

$1 m = 1000 mm$

$\rho = 5.23 * 10^{-6} \frac{kg}{(1*10^-3 m)^3}$

$\rho = 5.23 * 10^{-6} \frac{1*10^9 kg}{m^3}$

$\rho = 5.23 * 10^3 kg/m^3$

$\rho = 5230 kg/m^3$

8 0

3 years ago

Exposure to a sufficient quantity of ultraviolet will redden the skin, producing erythema - a sunburn. The amount of exposure ne

ivann1987 [24]

Answer:

Energy = 7.83 x 10⁻¹⁹ J

Energy = 6.63 x 10⁻¹⁹ J

Explanation:

The energy of a photon in terms of wavelength can be calculated by the following formula:

$Energy = \frac{hc}{\lambda}\\$

where,

h = Plank's Constant = 6.63 x 10⁻³⁴ Js

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light

Now, for λ = 254 nm = 2.54 x 10⁻⁷ m:

$Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{2.54\ x\ 10^{-7}\ m}\\$

Energy = 7.83 x 10⁻¹⁹ J

Now, for λ = 300 nm = 3 x 10⁻⁷ m:

$Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{3\ x\ 10^{-7}\ m}\\$

Energy = 6.63 x 10⁻¹⁹ J

7 0

2 years ago

A bicyclist steadily speeds up from rest to 11.0m/s during a 5.50s time interval. Determine all unknowns and answer the followin

Marina86 [1]

Explanation:

acceleration is 2 m/s^2

v-u/t

distance travelled is 30.25 meter

(v^2-u^2)/2a

11*11/2*2

121/4

30.25 m

5 0

2 years ago

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case

Murljashka [212]

Answer:

The answer is below

Explanation:

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

cable for the following cases:

a. The load moves downward at a constant velocity

b. The load accelerates downward at a rate 0.4 m/s??

C. The load accelerates upward at a rate 0.4 m/s??

Solution:

Acceleration due to gravity (g) = 10 m/s²

a) Given that the mass of the crane (m) is 140 kg. If the load moves downward, the tension (T) is given by:

mg - T = ma

Since the load has a constant velocity, hence acceleration (a) = 0. Therefore:

mg - T = m(0)

mg - T = 0

T = mg

T = 140(10) = 1400 N

T = 1400 N

b) If the load moves downward, the tension (T) is given by:

mg - T = ma

T = mg - ma = m(g - a)

T = 140(10 - 0.4) = 140(9.96) = 134.4

T = 134.4 N

c) If the load moves upward, the tension (T) is given by:

T - mg = ma

T = ma + mg = m(a + g)

T = 140(0.4 + 10) = 140(10.4)

T = 145.6 N

2) To find the distance (s) if the load move from rest (u= 0) and accelerates for 20 seconds (t = 20). We use:

s = ut + (1/2)gt²

s = 0(20) + (1/2)(10)(20)²

s = 2000 m

7 0

2 years ago