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2 years ago

Helppppppp pleaseeee

Physics

1 answer:

SVETLANKA909090 [29]2 years ago

3 0

Answer: (Sorry, but I don't know how to calculate mass)

1. 15 N

2. 0.4921 $\frac{ft}{s^2}$ (feet per second squared)

4. 150 N

5. 8.202 feet per second squared

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A truck loaded with sand accelerates along a highway. The driving force on the truck remains constant. What happens to the accel

pickupchik [31]

Answer:

Acceleration will increase.

Explanation:

The relation between force, mass and acceleration according to the Newton's second law of motion is given as:

F = ma

We are given that the driving force on the truck remains constant, so F is constant here. We can rewrite the above equation as:

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

Since, F is constant, the acceleration of the truck is inversely proportional to the mass.

There is a hole at the bottom of the truck through which the sand is being lost at a constant rate. Since, the sand is being lost, the overall mass of the truck is being reduced.

Since, the acceleration of the truck is inversely proportional to the mass, the reduced mass will result in an increased acceleration.

So, the acceleration of the truck will increase.

4 0

3 years ago

The volume flow rate of the water supplied by a well is 2.0×10−4m3/s.The well is 40.0 m deep. (a) What is the power output of th

MaRussiya [10]

Answer:

a). $P=78.4W$

b). $P=392kPa$

c.) It must be at the bottom

Explanation:

Given:

Volume flow $V_f=2.0x10^{-4}m^3/s$

Well depp $h=40.m$

The power output of the pum

$W=F*d$

$F=m*g$

$m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg$

$W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2$

$W=78.4J$

$P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W$

The pressure of difference the pum

Δ $P=p*g*y'$

Δ $P=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa$

$P=392kPa$

It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way

4 0

3 years ago

Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid

NeTakaya

Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

The mass of block A = 3·m

The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ = 4·m·v₃

$\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0$

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

$\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2$

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

$m \cdot g \cdot h = \dfrac{9}{8} \cdot m\cdot v_0^2$

$h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g } = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g } = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot v_0^2 \approx 0.1147959 \cdot v_0^2$

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

7 0

2 years ago

How much force is required to move a tractor 60m, when using 250J of work?

lianna [129]

Answer:

Work done,W= 250J

Displacement , s = 60

We know that, Work done = Force x displacement

i.e , W = Fxs

250 J = F x 60m

F = 250/60

=4.16 N

Hence , 4.16 N of Force is applied on the body.

5 0

2 years ago

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What is the largest component of M1?

mariarad [96]

The largest component of M1 is- Currency.

8 0

3 years ago