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Deffense [45]
2 years ago
8

Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s and a speed of 42.0 m/s.

Physics
1 answer:
ss7ja [257]2 years ago
8 0

Answer:

Force, |F| = 2100 N

Explanation:

It is given that,

Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s, \dfrac{m}{t}=50\ kg/s

Initial speed, v = 42 m/s

The momentum is reduced to zero, final speed, v = 0

The relation between the force and the momentum is given by :

F=\dfrac{p}{t}

F=\dfrac{mv}{t}

F=50\ kg/s\times 42\ m/s

|F| = 2100 N

So, the magnitude of the force exerted on the wall is 2100 N. Hence, this is the required solution.

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Four ways to encourage immigrants positively
Svetach [21]

Answer:

See below.

Explanation:

1) Build relationships with key leaders and organizations

Early on, communicate with immigrant community leaders and groups to establish contacts, learn about these communities, and express your desire to include immigrant citizens in civic and political life. Make a list of them and keep it updated. Individuals and groups, be on the lookout for opportunities to meet with them and participate in local activities. Keep in touch with them on a regular basis and attend their events and activities.

2) Identify issues that immigrants care about

Identify problems of interest to immigrant populations through personal interactions, surveys, and meetings with local groups, and be prepared to include these themes in community talks and other public engagement initiatives. If a need is found, you can start with conversations and civic involvement in immigrant communities, since this can create channels for participation and leadership development for people who might not otherwise participate. Demonstrate how civic involvement may assist immigrants in achieving their goals and aspirations.

3) Overcome language barriers

Every participant should be prepared to participate, understood, and comprehend what others are saying as part of public engagement activities. Translation equipment and services should be provided, as well as outreach and issue background materials that are relevant for your community. The translation services should be mentioned in outreach for public engagement activities.

Ensure that documents are translated by native speakers or those who are entirely proficient in order for translations to be comprehended by readers. When conducting polls and surveys, it is preferable to ask questions in the native language of the residents. You may contact people with varied literacy levels through face-to-face, radio, and other non-written interactions.

4) Use effective media and outstretch strategies

Integrate immigrant-focused local and regional print and electronic media into your communication plan. Build ties with these media outlets, provide them access to information and local officials, send them news, announcements, and job ads, and work with them as partners to develop effective outreach to increase public participation. Inform suitable community, service, or business groups, schools, congregations, and others about a forthcoming public engagement activity. Request that the leaders of these organizations include particular solicitations for their immigrant members, as well as follow up with and assist those who are interested in attending.

6 0
2 years ago
lood flows through a section of a horizontal artery that is partially blocked by a deposit along the artery wall. As a hemoglobi
uysha [10]

Answer:

P_1 - P_2 =219.62\ Pa

Explanation:

given,

density of blood = 1060 kg/m³

v₂ = 0.800 m/s

v₁ = 0.475 m/s

change in pressure calculation

using Bernoulli's equation

P_1 +\dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2

both are at same level

P_1 - P_2 =\dfrac{1}{2}\rho (v_2^2- v_1^2)

P_1 - P_2 =\dfrac{1}{2}1060\times (0.8^2- 0.475^2)

P_1 - P_2 =219.62\ Pa

the change in pressure is equal to 219.62 Pa

5 0
3 years ago
If two lines in a system of linear equations have the same slope and same y-intercept, how many solutions will the system have?
arsen [322]

Answer:

We will have <u>infinite solutions </u>to the system of linear equations.

Explanation:

Well, when we have two lines with the <u>same slopes and the same y-interception</u>, both of them <u>are overlapped, </u>so we will have <u>infinite solutions </u>to the system of linear equations.

This kind of system is called <u>dependent system.</u>

I hope it helps you!

8 0
3 years ago
Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 54 km/h. After four hours
MariettaO [177]
With that information you can only suppose a uniformly accelerated motion.  This is, acceleration is constant.

Then, acceleration = change in velocity / change in time = (58 -54)km/h / 2 h = 4km/h / 2 h = 2 km/h^2

Then the equation for velocity, V is

V = Vo + a*t = Vo + 2 (km/h^2)  * t = Vo +  2t

Vo is the initial velocity, which you can find using V = 54km/h and t = -2

Vo = V after 2 hours - a*(2hours) = 54km/h - 2(km/h^2)*2h = 54km/k - 4km/h = 50km/h

Then, the equation is: V = 50 km/h + 2t

Valid for constant acceleration. 
5 0
2 years ago
MATHPHYS CAN U HELP ME PLEASE
ludmilkaskok [199]

Explanation:

(1) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.041 kg) (2090 J/kg/°C) (0°C − (-11°C)) = 942.59 J

The heat added to melt the ice is:

q = mL = (0.041 kg) (3.33×10⁵ J/kg) = 13,653 J

The heat added to warm the water to 100°C is:

q = mCΔT = (0.041 kg) (4186 J/kg/°C) (100°C − 0°C) = 17,162.6 J

The heat added to evaporate the water is:

q = mL = (0.041 kg) (2.26×10⁶ J/kg) = 92,660 J

The heat added to warm the steam to 115°C is:

q = mCΔT = (0.041 kg) (2010 J/kg/°C) (115°C − 100°C) = 1236.15 J

The total heat needed is:

q = 942.59 J + 13,653 J + 17,162.6 J + 92,660 J + 1236.15 J

q = 125,654.34 J

(2) When the first two are mixed:

m C₁ (T₁ − T) + m C₂ (T₂ − T) = 0

C₁ (T₁ − T) + C₂ (T₂ − T) = 0

C₁ (6 − 11) + C₂ (25 − 11) = 0

-5 C₁ + 14 C₂ = 0

C₁ = 2.8 C₂

When the second and third are mixed:

m C₂ (T₂ − T) + m C₃ (T₃ − T) = 0

C₂ (T₂ − T) + C₃ (T₃ − T) = 0

C₂ (25 − 33) + C₃ (37 − 33) = 0

-8 C₂ + 4 C₃ = 0

C₂ = 0.5 C₃

Substituting:

C₁ = 2.8 (0.5 C₃)

C₁ = 1.4 C₃

When the first and third are mixed:

m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0

C₁ (T₁ − T) + C₃ (T₃ − T) = 0

(1.4 C₃) (6 − T) + C₃ (37 − T) = 0

(1.4) (6 − T) + 37 − T = 0

8.4 − 1.4T + 37 − T = 0

2.4T = 45.4

T = 18.9°C

(3) Heat gained by the ice = heat lost by the tea

mL + mCΔT = -mCΔT

m (3.33×10⁵ J/kg) + m (2090 J/kg/°C) (30.8°C − 0°C) = -(0.176 kg) (4186 J/kg/°C) (30.8°C − 32.8°C)

m (397372 J/kg) = 1473.472 J

m = 0.004 kg

m = 4 g

4 grams of ice is melted and warmed to the final temperature, which leaves 128 grams unmelted.

(4) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.028 kg) (2090 J/kg/°C) (0°C − (-67°C)) = 3920.84 J

The heat added to melt the ice is:

q = mL = (0.028 kg) (3.33×10⁵ J/kg) = 9324 J

The heat added to warm the melted ice to T is:

q = mCΔT = (0.028 kg) (4186 J/kg/°C) (T − 0°C) = (117.208 J/°C) T

The heat removed to cool the water to T is:

q = -mCΔT = -(0.505 kg) (4186 J/kg/°C) (T − 27°C)

q = (2113.93 J/°C) (27°C − T) = 57076.11 J − (2113.93 J/°C) T

The heat removed to cool the copper to T is:

q = -mCΔT = -(0.092 kg) (387 J/kg/°C) (T − 27°C)

q = (35.604 J/°C) (27°C − T) = 961.308 J − (35.604 J/°C) T

Therefore:

3920.84 J + 9324 J + (117.208 J/°C) T = 57076.11 J − (2113.93 J/°C) T + 961.308 J − (35.604 J/°C) T

13244.84 J + (117.208 J/°C) T = 58037.418 J − (2149.534 J/°C) T

(2266.742 J/°C) T = 44792.58 J

T = 19.8°C

(5) Kinetic energy of the hammer = heat absorbed by ice

KE = q

½ mv² = mL

½ (0.8 kg) (0.9 m/s)² = m (80 cal/g × 4.186 J/cal × 1000 g/kg)

m = 9.68×10⁻⁷ kg

m = 9.68×10⁻⁴ g

(6) Heat rate = thermal conductivity × area × temperature difference / thickness

q' = kAΔT / t

q' = (1.09 W/m/°C) (4.5 m × 9 m) (10°C − 4°C) / (0.09 m)

q' = 2943 W

After 10.7 hours, the amount of heat transferred is:

q = (2943 J/s) (10.7 h × 3600 s/h)

q = 1.13×10⁸ J

q = 113 MJ

6 0
3 years ago
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